Acid/Base Equilibria
HCl: acid NaOH: base Fe(OH)3: base H2SO4: acid Ca(OH)2: base HNO3: acid
Identify whether or not the following are an Arrhenius acid or an Arrhenius base: -HCl -NaOH -Fe(OH)3 -H2SO4 -Ca(OH)2 -HNO3
pKa = -log[Ka] = -log(1.8 x 10^-5) = 4.74 pKa = 4.74
If the Ka of an acid is 1.8 x 10^-5, then what is its pKa?
pH or pOH
The Henderson Hasselbalch equation is used to estimate the ________________ of a buffer solution.
Lower Higher
The ____________ the pKa, the stronger the acid. The ____________ the pKb, the stronger the base.
The bicarbonate buffer system.
The reaction shown is an example of what?
Polyvalent acid Polyvalent base
A _______________ acid is an acid that is capable of donating more than one acid equivalent (H3O+) A _______________ base is a base that is capable of donating more than one base equivalent (OH-) or accepting more than one proton.
Amphoteric species
A _______________ species is a substance that can behave as an acid or base depending on the reaction it is in.
Amphiprotic species
A _______________ species is an amphoteric species that can specifically act as either a Bronsted-Lowry acid or Bronsted-Lowry base.
Answer: C. A higher Ka implies a stronger acid. Weak acids usually have a Ka that is several orders of magnitude below 1. The pKa of a compound is the pH at which there are equal concentrations of acid and conjugate base; the pKa of this compound would be -log(1) = 0. With such a low pKa, this compound must be an acid. Therefore, the pH of any concentration of this compound must be below 7.
A solution is prepared with an unknown concentration of a theoretical compound with a Ka of exactly 1.0. What is the pH of this solution? A. Higher than 7 B. Exactly 7 C. Less than 7 D. There is not enough information to answer the question
Less than Greater than
A solution would be acidic if it had a pH that is ________ than 7. A solution would be basic if it had a pH that is _______________ than 7
Ka * Kb = Kw Ka * (3.7 x 10^-4) = (1 x 10^-14) Ka = (1 x 10^-14)/(3.7 x 10^-4) = (2.7 x 10^-11) Ka = 2.7 x 10^-11 pKa: pKa = -log[Ka] = -log[2.7 x 10^-11] = 10.57 pKa = 10.57 pKb: pKb = -log[Kb] = -log[3.7 x 10^-4] = 3.43 or... pKa + pKb = 14 pKb = 14 - 10.57 = 3.43 pKb = 3.43
Calculate the Ka value for the methylammonium ion (CH3NH3+). Kb for methylamine (CH3NH2) is 3.7 x 10^-4 What is the pKa value of CH3NH3+? What is the pKb value of CH3NH2?
CH3COOH + H2O <-------> CH3COO- + H3O+ I 2.0 0 0 C -x +x +x E 2.0 - x x x Ka = [CH3COO-][H3O+]/[CH3COOH] 1.8 x 10^-5 = [x][x]/[2.0 - x] (assume x << 1) 1.8 x 10^-5 = x^2/2.0 2.0 * 1.8 x 10^-5 = x^2 = 3.6 x 10^-5 sqrt(3.6 x 10^-5) = x x = 6 x 10^-3 Ka = 6 x 10^-3
Calculate the concentration of H3O+ in a 2.0 M aqueous solution of acetic acid, CH3COOH. (Ka = 1.8 x 10^-5)
Since NaOH is a strong base, it will dissociate at 100%. NaOH -----> Na+ + OH- Since ionization is 100%, [NaOH] = [OH-] [OH-] = 0.20 M pOH = -log[OH-] = -log(0.20) = 0.70 pOH + pH = 14 pH = 14 - pOH = 14 - 0.70 = 13.30 pH = 13.30
Calculate the pH of a 0.20 M solution of NaOH.
[NH3] = 0.500 M NH3 + H2O <-------> NH4+ + OH- I 0.500 0 0 C -x +x +x E 0.500 - x x x Kb = [NH4+][OH-]/[NH3] 1.8 x 10^-5 = [x][x]/[0.500 - x] assume x is negligible 1.8 x 10^-5 = x^2/0.500 (1.8 x 10^-5) * 0.500 = x^2 9.0 x 10^-6 = x^2 sqrt(9.0 x 10^-6) = x = 3 x 10^-3 x = [OH-] = 3 x 10^-3 pOH = -log[OH-] = -log(3 x 10^-3) = 2.52 pH + pOH = 14 pH = 14 - 2.52 pH = 11.48
Calculate the pH of a 0.500 M solution of NH3. Kb = 1.8 x 10^-5
Lewis Acid: HCl Lewis Base: H2O
In the following equation, identify which substance is: -Lewis Acid -Lewis Base
[H3O+] = 2.2 x 10^-3 M [H3O+][OH-] = Kw [2.2 x 10^-3][OH-] = 1.0 x 10^-14 [OH-] = (1.0 x 10^-14)/(2.2 x 10^-3) = 4.5 x 10^-12 [OH-] < [H3O+] = acidic [OH-] = 4.5 x 10^-12 and the solution is acidic.
Lemon juice has a concentration of H3O+ ions at 2.2 x 10^-3 M. Find the concentration of OH- ions in the solution, and classify whether the solution is acidic or basic.
-Perchloric acid -Perchlorate
Name the acid and it's anion: -HClO4 -ClO4-
-Hydrofluoric acid -Fluoride
Name the acid and it's anion: -HF -F-
-Nitrous acid -Nitrite
Name the acid and it's anion: -HNO2 -NO2-
-Nitric acid -Nitrate
Name the acid and it's anion: -HNO3 -NO3-
The one on the left is stronger since the electronegative atom (Cl) is closer to the red acidic proton in the left molecule than the right molecule.
Of the two acidic molecules, which one is stronger? Explain your reasoning.
Mass
The gram equivalent weight of an acid/base is the ______________ of the acid/base that produces 1 acid/base equivalent.
Stronger Stronger
The higher the Ka, the ____________ the acid. The higher the Kb, the _____________ the base.
pH = -log[H3O+] [H3O+] = 1.0 x 10^-7 pH = -log[1.0 x 10^-7] = 7 pH = 7.00
What is the pH of water?
14.
pKa + pKb = _________
Polyprotic species
A ______________ species is a molecule that is capable of donating more than one proton.
We're including the 0.15 M NH3 along with the 0.35 M NH4NO3. NH3 + H2O <----------> NH4+ + OH- I 0.15 0.35 0 C -x +x +x E 0.15 - x 0.35 + x x Kb = [NH4+][OH-]/[NH3] 1.8 x 10^-5 = [0.35 + x][x]/[0.15 - x] (assume x << 1) 1.8 x 10^-5 = 0.35 * x/0.15 (0.15 * 1.8 x 10^-5)/0.35 = x x = 7.7 x 10^-6 x = [OH-] pOH = -log[OH-] = -log[7.7 x 10^-6] = 5.11 pH + pOH = 14 pH = 14 - 5.11 = 8.89 pH = 8.89
Calculate the pH of a solution that is 0.15 M NH3 (Kb = 1.8 x 10^-5) and 0.35 M NH4NO3.
Find moles: 0.52 g/ 40 g/mol = 0.013 mol Find molarity: 0.013 mol/ 0.431 L = 0.03 M Find pH pOH = -log[0.03] = 1.52 14 - 1.52 = 12.48 pH = 12.48
Calculate the pH of an aqueous solution that contains 0.52 g of NaOH in a total volume of 431 mL. (molar mass of NaOH is 40 g/mol)
Find moles: (0.56 g/ 36 g/mol) = 0.0156 mol Find molarity: 0.0156 mol/ 0.576 L = 0.027 M Find pH: -log[0.027] = 1.57 pH = 1.57
Calculate the pH of an aqueous solution that contains 0.56 g of HCl in a total volume of 576 mL. (molar mass of HCl is 36 g/mol)
Bronsted-Lowry Acid & Base: -Bronsted-Lowry Acid is the proton donor -Bronsted-Lowry Base is the proton acceptor Conjugate Acid & Base: -Conjugate Acid is formed from a Bronsted-Lowry Base accepting a proton -Conjugate Base is formed from a Bronsted-Lowry Acid donating a proton Lewis Acid & Base: -Lewis Acid is an electron pair acceptor -Lewis Base is an electron pair donor Arrhenius Acid & Base: -Acid produces an excess of H+ ions when it dissociates in an aqueous solution -Base produces an excess of OH- ions when it dissociates in an aqueous solution
Describe the differences between the following pairs: -Bronsted-Lowry Acid & Bronsted-Lowry Base -Conjugate Acid & Conjugate Base -Lewis Acid & Lewis Base -Arrhenius Acid & Arrhenius Base
Weak acids
Examples of ___________ ____________ includes the following: -HF (hydrofluoric acid) -CH3COOH (acetic acid) -CH3OH (methanol)
Weak bases
Examples of ____________ ____________ includes the following: -NH3 (ammonia) -C6H5NH2 (aniline)
Polyvalent bases
Examples of _____________ ________________ includes the following: -Al(OH)3 (aluminum hydroxide) -Ca(OH)2 (calcium hydroxide) -Mg(OH)2 (magnesium hydroxide)
Polyvalent acids
Examples of _______________ ______________ includes the following: -H2SO4 (sulfuric acid) -H3PO4 (phosphoric acid) -H2CO3 (carbonic acid)
By increasing the number of moles of the buffer in comparison to the moles of the strong acid/base.
How can the effectiveness of a buffer be strengthened?
Conjugate base Conjugate acid
The stronger the acid, the weaker the _______________ will be. The stronger the base, the weaker the ______________ will be.
-NaOH ---> Na+ -CaOH ---> Ca+ -NH3 ---> NH4+ -Mg(OH)2 ---> MgOH+ -NH4OH ---> NH4+ -KOH ---> K+
What is the conjugate acid of the following bases? -NaOH -CaOH -NH3 -Mg(OH)2 -NH4OH -KOH
-HCl ---> Cl- -CH3COOH ---> CH3COO- -H2SO4 ---> HSO4- -CH3OH ---> CH3O- -HI ---> I- -HNO3 ---> NO3- -HClO4 ---> ClO4-
What is the conjugate base of the following acids? -HCl -CH3COOH -H2SO4 -CH3OH -HI -HNO3 -HClO4
Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) NH3 = A- NH4+ = HA pKa = -log[Ka] find Ka using Kb Ka * Kb = Kw Ka * 1.8 x 10^-5 = 1.0 x 10^-14 Ka = 1.0 x 10^-14/1.8 x 10^-5 = 5.6 x 10^-10 pKa = -log[5.6 x 10^-10] = 9.25 pH = 9.25 + log(0.24/0.20) pH = 9.25 + log(1.20) = 9.25 + 0.080 = 9.33 pH = 9.33 pH = 9.25 + log9
What is the pH of a buffer solution that is 0.24 M NH3 and 0.20 M NH4+Cl-? (Kb of NH3 = 1.8 x 10^-5)
Normality = N Volume = V CH3COOH: M(i) * V(i) = M(f) * V(f) M(f) = M(i) * V(i)/V(f) V(f) = 1 L + 500 mL M(f) = (0.05 M * 1 L)/1.5 L = 0.033 M CH3COO-: M(i) * V(i) = M(f) * V(f) M(f) = M(i) * V(i)/V(f) M(f) = (1 M * 0.5 L)/1.5 L = 0.33 M Use Henderson-Hasselbach equation: pH = pKa + log[A-]/[HA] pKa = -log[Ka] = -log[1.8 x 10^-5] = 4.74 pH = 4.74 + log(0.33/0.033) = 4.74 + 1 = 5.74 pH = 5.74
What is the pH of a solution made from 1 L of 0.05 M acetic acid (CH3COOH, Ka = 1.8 x 10^-5) mixed with 500 mL of 1 M acetate (CH3COO-)?
Ka of CH3OH = 2.9 x 10^-16 pKa = -log[Ka] = -log[2.9 x 10^-16] = 15.54 pKa = 15.54
What is the pKa of CH2OH? Ka = 2.9 x 10^-16
-Acetic acid -Acetate
Name the acid and it's anion: -CH3COOH -CH3COO-
-Carbonic acid -Carbonate
Name the acid and it's anion: -H2CO3 -CO3(2-)
-Sulfuric acid -Sulfate
Name the acid and it's anion: -H2SO4 -SO4(2-)
-Boric acid -Borate
Name the acid and it's anion: -H3BO3 -BO3(3-)
-Phosphoric acid -Phosphate
Name the acid and it's anion: -H3PO4 -PO4(3-)
acid-base neutralization reaction
A ______________________________ reaction is a reaction between an acid and a base to form a solution of a salt and water.
10:1 1:10
A buffer is considered to be effective when the HA/A- ratio ranges anywhere between __________ or _____________
Greater Less
A solution would be considered acidic if the concentration of hydronium ions is _____________ than the concentration of hydroxide ions. A solution would be considered basic if the concentration of hydronium ions is _____________ than the concentration of hydroxide ions.
H3O+ OH-
An equivalent for an acid would be ___________ while an equivalent for a base would be ____________.
Al(OH)3: Al(OH)3 is a base, and has 3 OH- groups in it. Thus has 3 base equivalents. The example also mentions that there is 2 M of Al(OH)3. 3 * 2 = 6 N Al(OH)3 = 6 H2SO4: H2SO4 is an acid, and has 2 H+ groups to make 2 H3O+ ions. Thus has 2 base equivalents. The example also mentions that there is 16 M of H2SO4. 16 * 2 = 32 N H2SO4 = 32
Calculate the normality of the following solutions: 2 M Al(OH)3 16 M H2SO4
Remember: Na+ cations will not react with water, but CH3COO- anions will, since this salt is formed from a weak acid and a strong base. So make a chemical equation for CH3COO- and water. CH3COO- + H2O <------> CH3COOH + OH- I 0.25 0 0 C -x +x +x E 0.25 - x x x Kb = [CH3COOH][OH-]/[CH3COO-] Kb = [x][x]/[0.25 - x] (assume x<<1) Don't know the Kb of CH3COO-, so we find it by using the Ka of acetic acid. Ka * Kb = Kw 1.8 x 10^-5 * Kb = 1.0 x 10^-14 Kb = 1.0 x 10^-14/1.8 x 10^-5 = 5.6 x 10^-10 5.6 x 10^-10 = x^2/0.25 0.25 * 5.6 x 10^-10 = x^2 1.4 x 10^-10 = x^2 sqrt(1.4 x 10^-10) = x x = 1.2 x 10^-5 = [OH-] pOH = -log[OH-] = -log[1.2 x 10^-5] = 4.92 pH + pOH = 14 pH = 14 - 4.92 = 9.08 pH = 9.08 Solution is basic
Calculate the pH of 0.25 M solution of CH3COO-Na+ and state whether the solution is acidic, basic, or neutral. Ka of CH3COOH = 1.8 x 10^-5
[CH3COOH] = 1.00 M CH3COOH + H2O <------> CH3COO- + H3O+ Initial 1.00 M 0 0 Change -x +x +x Equilibrium 1.00 - x x x Find Ka: Ka = products/reactants = [x][x]/[1.00-x] assume x << 1, makes x negligible in 1.00 - x 1.8 x 10^-5 = x^2/1.00 1.8 x 10^-5 = x^2 sqrt(1.8 x 10^-5) = x x = 0.0042 x = [H3O+] = 0.0042 pH = -log[H3O+] = -log[0.0042] = 2.38 pH = 2.38
Calculate the pH of 1.00 M solution of vinegar (acetic acid in water). Ka = 1.8 x 10^-5
Nitric acid (HNO3) is a strong acid. Strong acids undergoes 100% ionization in a solution. HNO3 + H2O 100%-----> H3O+ + NO3- Since ionization is 100%, [HNO3] = [H3O+] [H3O+] = 0.030 M pH = -log[H3O+] = -log(0.030) pH = 1.52 pH = 1.52
Calculate the pH of a 0.030 M HNO3 solution.
Remember: Cl- anions will not affect pH, NH4+ cations will. So make a chemical equation for NH4+ and water. NH4+ + H2O <---------> NH3 + H3O+ I 0.050 0 0 C -x +x +x E 0.050 - x x x Ka = [x][x]/[0.050 - x] (assume x << 1) We don't know the Ka of NH4+, so we use the Kb of the conjugate base NH3 to find Ka. Ka * Kb = Kw Ka * 1.8 x 10^-5 = 1.0 x 10^-14 Ka = 1.0 x 10^-14/1.8 x 10^-5 = 5.6 x 10^-10 5.6 x 10^-10 = x^2/0.050 0.050 * 5.6 x 10^-10 = x^2 2.8 x 10^-11 = x^2 sqrt(2.8 x 10^-11) = x = 5.3 x 10^-6 x = [H3O+] pH = -log[H3O+] = -log[5.3 x 10^-6] = 5.28 pH = 5.28 Solution is acidic
Calculate the pH of a 0.050 M solution of NH4+Cl- and state whether the solution is acidic, basic, or neutral. Kb of NH3 = 1.8 x 10^-5
We're including the 1.00 M of CH3COONa along with the 1.00 M of CH3COOH. CH3COOH + H2O <--------> CH3COO- + H3O+ I 1.00 1.00 0 C -x +x +x E 1.00 - x 1.00 + x x Ka = [H3O+][CH3COO-]/[CH3COOH] 1.8 x 10^-5 = [x][1.00 + x]/[1.00 - x] (assume x << 1) 1.8 x 10^-5 = x * 1.00/1.00 x = 1.8 x 10^-5 x = [H3O+] pH = -log[H3O+] = -log[1.8 x 10^-5] = 4.74 pH = 4.74
Calculate the pH of a solution that is 1.00 M of CH3COOH (Ka = 1.8 x 10^-5) and 1.00 M of NaOH.
Find moles: mol = g/molar mass mol = 0.11 g/74 g/mol = 0.0015 mol Multiply by 2, since there are two moles of OH- in Ca(OH)2. 0.0015 x 2 = 0.0030 mol Find concentration (molarity): molarity = moles/liters convert: 250 mL/1000 = 0.250 L molarity = 0.0030 mol/0.250 L = 0.012 M [Ca(OH)2] = 0.012 M Ca(OH)2 -----> Ca(2+) + 2OH- Since 100% ionization, [Ca(OH)2] = [OH-] [OH-] = 0.012 M pOH = -log[OH-] = -log(0.012) = 1.92 pH + pOH = 14 pH = 14 - 1.92 = 12.08 pH = 12.08
Calculate the pH of an aqueous solution that contains 0.11 g of Ca(OH)2 in a total volume of 250 mL. (molar mass of Ca(OH)2 is 74 g/mol)
Find moles: 0.62 g/ 171 g/mol = 0.0036 Ba(OH)2 has 2 acid equivalents 0.0036 x 2 = 0.0072 mol Find molarity: 0.0072 mol/ 0.35 L = 0.021 M Find pH: pOH = -log[0.021] = 1.68 14 - 1.68 = 12.32 pH = 12.32
Calculate the pH of an aqueous solution that contains 0.62 g of Ba(OH)2 in a total volume of 350 mL. (molar mass of Ba(OH)2 is 171 g/mol)
Find moles: (0.3 g/ 98 g/mol) = 0.007 H2SO4 has 2 acid equivalents 0.007 x 2 = 0.014 mol Find molarity: 0.014 mol/ 0.051 L = 0.275 M Find pH: -log [0.275] = 0.56 pH = 0.56
Calculate the pH of an aqueous solution that contains 0.7 g of H2SO4 in a total volume of 51 mL. (molar mass of H2SO4 is 98 g/mol)
Increases Stronger
Electronegative elements that are positioned closer to an acidic proton _____________ the acid strength by pulling electron density out of the bond holding the acidic proton. Furthermore, molecules with electronegative elements are ____________ in acid strength than molecules without electronegative elements.
Strong bases
Examples of __________ ___________ includes the following: -NaOH (sodium hydroxide) -KOH (potassium hydroxide) -Ca(OH)2 (calcium hydroxide) -Mg(OH2) (magnesium hydroxide)
Strong acids
Examples of __________ ____________ includes the following: -HClO4 (perchloric acid) -HCl (hydrochloric acid) -HBr (hydrobromic acid) -HI (hydroiodic acid) -H2SO4 (sufuric acid) -HNO3 (nitric acid)
An Arrhenius acid can be identified because their formula begins with an H. An Arrhenius base can be identified because their formula ends with an OH.
How can you identify whether or not an acid or base is an Arrhenius one?
Find concentration of NaOH 0.005 mol NaOH/0.50 L buffer = 0.01 M NaOH Since NaOH is a strong base, it is the same concentration as OH NaOH will react with NH4+ (strong bases will react with acids) A- = NH3 HA = NH4+ NH4+ + OH- <---------> H2O + NH3 I 0.20 0.01 0.24 C -0.01 -0.01 +0.01 E 0.19 0 0.25 pH = pKa + log [A-]/[HA] find Ka with Kb Ka * Kb = Kw Ka * 1.8 x 10^-5 = 1.0 x 10^-14 Ka = 1.0 x 10^-14/1.8 x 10^-5 Ka = 5.6 x 10^-10 pKa = -log[Ka] = -log[5.6 x 10^-10] = 9.25 pH = 9.25 + log(0.25/0.19) pH = 9.25 + log(1.31) = 9.25 + 0.12 = 9.37 pH = 9.37 The pH of the buffer slightly increased with the addition of NaOH, but the change is not substantial.
If 0.005 mol of NaOH is added to 0.50 L of buffer solution (0.24 M NH3 + 0.20 M NH4+Cl-), what is the resulting pH? If the original buffer had a pH = 9.33 how much of a change is present when NaOH was added? (Kb of NH3 = 1.8 x 10^-5)
Since HCl is a strong acid, [HCl] = [H3O+] Find concentration of HCl 0.03 mol/0.50 L = 0.06 M HCl Base reacts to strong acids, so HCl reacts with NH3 A- = NH3 HA = NH4+ NH3 + H3O+ <---------> NH4+ + H2O I 0.24 0.06 0.20 C - 0.06 - 0.06 + 0.06 E 0.18 0 0.26 pH = pKa + log[A-]/[HA] Find Ka with Kb Ka * Kb = Kw Ka * (1.8 x 10^-5) = 1.0 x 10^-14 Ka = 1.0 x 10^-14/1.8 x 10^-5 Ka = 5.6 x 10^-10 pKa = -log[Ka] = -log[5.6 x 10^-10] = 9.25 pKa = 9.25 pH = 9.25 + log(0.18/0.26) = 9.25 + log(0.69) pH = 9.25 - 0.16 = 9.09 pH = 9.09 The pH of the buffer slightly decreased with the addition of HCl, but the change is not substantial.
If 0.03 mol of HCl is added to 0.50 L of the buffer solution (0.24 M NH3 and 0.20 M NH4+Cl-), what is the resulting pH? If the original buffer had a pH = 9.33 how much of a change is present when HCl was added? (Kb of NH3 = 1.8 x 10^-5)
Stop working Denature Disable
If our blood reaches a pH below physiological pH, the cells of our body will _______ _________________. Enzymes will _________________, which will ___________ their catalytic abilities
14
In acid-base calculations: pH + pOH = _________
[H3O+] [OH-]
In acid-base calculations: pH = -log_______ pOH = -log_______
Bronsted-Lowry Acid: H2O Bronsted-Lowry Base: HCO3- Conjugate Acid: H2CO3 Conjugate Base: OH-
In the following equation, identify which substance is: -Bronsted-Lowry Acid -Bronsted-Lowry Base -Conjugate Acid -Conjugate Base
Bronsted-Lowry Acid: H2O Bronsted-Lowry Base: NH3 Conjugate Acid: NH4+ Conjugate Base: OH-
In the following equation, identify which substance is: -Bronsted-Lowry Acid -Bronsted-Lowry Base -Conjugate Acid -Conjugate Base
Bronsted-Lowry Acid: H2SO4 Bronsted-Lowry Base: H2O Conjugate Acid: H3O+ Conjugate Base: HSO4-
In the following equation, identify which substance is: -Bronsted-Lowry Acid -Bronsted-Lowry Base -Conjugate Acid -Conjugate Base
The hydrogen chloride molecule (hydrochloric acid) is the Bronsted-Lowry Acid, since it donates a H+ proton to water. The water molecule is the Bronsted-Lowry Base, since it accepts a H+ proton from the hydrochloric acid. The conjugate acid is the hydronium (H3O+) ion, since it was formed as a result of water accepting the H+ proton. The conjugate base is the chloride (Cl-) ion, since it was formed as a result of HCl donating a proton to water.
In the following equation, identify which substance is: -Bronsted-Lowry Acid -Bronsted-Lowry Base -Conjugate Acid -Conjugate Base
Lewis Acid: Ag+ Lewis Base: 2NH3
In the following equation, identify which substance is: -Lewis Acid -Lewis Base
Lewis Acid: BF3, since it is accepting an electron pair Lewis Base: F-, since it is donating an electron pair
In the following equation, identify which substance is: -Lewis Acid -Lewis Base
Lewis Acid: SO3 Lewis Base: O(2-)
In the following equation, identify which substance is: -Lewis Acid -Lewis Base
-Chromic acid -Chromate
Name the acid and it's anion: -H2CrO4 -CrO4(2-)
-Hydrobromic acid -Bromide
Name the acid and it's anion: -HBr -Br-
-Hydrochloric acid -Chloride
Name the acid and it's anion: -HCl -Cl-
-Hypochlorous acid -Hypochlorite
Name the acid and it's anion: -HClO -ClO-
-Chlorous acid -Chlorite
Name the acid and it's anion: -HClO2 -ClO2-
-Chloric acid -Chlorate
Name the acid and it's anion: -HClO3 -ClO3-
Strong acid + strong base Strong acid + weak base Weak acid + strong base Weak acid + weak base
The 4 kinds of neutralization reactions among acids and bases include: ____________ acid + ____________ base ___________ acid + _____________ base ___________ acid + _____________ base ___________ acid + ____________ base
pH = pKa + log([A-]/[HA]) pOH = pKb + log([B+]/[HB])
The Henderson Hasselbalch equation: pH = _______ + log(________/_________) pOH = _______ + log(_______/________)
Bicarbonate buffer system
The ________________ is a system in our blood that helps us maintain a physiological level of pH using carbonic acid and bicarbonate ions
Buffers
The following are common examples of ______________: - Acetic acid & sodium acetate CH3COOH + CH3COO-Na+ - Ammonia & ammonium chlorate NH3 + NH4+Cl-
Answer: B. The purpose of a buffer is to resist changes in the pH of a reaction. Buffers are not generally used to affect the kinetics of a reaction, so (C.) and (D.) are incorrect. (A.) is correct only in specific circumstances where the pH of the buffer solution itself is neutral. Many natural buffers systems maintain pH in acidic or basic ranges.
The function of a buffer is to: A. maintain a neutral pH B. resist changes in pH when small amounts of acid or bases are added C. slow down reactions between acids and bases D. speed up reactions between acids and bases
Water
The most common example of an amphiprotic amphoteric species is __________
An aqueous solution that resists changes in pH upon the addition of an acid or base. It is made either from a weak acid and its salt, which consists of the conjugate base and a cation, or from a weak base and its salt, which consists of the conjugate acid and an anion.
What is a buffer solution? How is a buffer solution made?
Answer: D. This question requires the application of the acid dissociation constant. Weak acids do not dissociate completely; therefore, all three species that appear in the balanced equation will be present in the solution. Hydrogen ions and conjugate base anions dissociate in equal amounts, so [H+] = [XO2-]. If the initial concentration of HXO2 was 2 M and some amount x dissociates, we will have x amount of H3O+ and XO2- at equilibrium, with 2 M - x amount of HXO2 at equilibrium. HXO2 + H2O <-----> H3O+ + XO2- 2M 0 0 -x +x +x 2M - x x x Ka = [H3O+][XO2-]/[HXO2] = [x][x]/[2-x] x<<1 Ka = x^2/2 3.2 x 10^-5 = x^2/2 x^2 = 6.4 x 10^-5 = 8 x 10^-3 M
What is the [H3O+] of a 2 M aqueous solution of a weak acid HXO2 with Ka = 3.2 x 10^-5? A. 6.4 x 10^-5 M B. 1.3 x 10^-4 M C. 4.0 x 10^-3 M D. 8.0 x 10^-3 M
Answer: C. NaOH is a strong base; as such, there will be 1.2 x 10^-5 M OH- in solution. Based on this information alone, the pOH must be between 4 and 5, and the pH must be between 9 and 10. pOH = -log[1.2 x 10^-5] = 4.92 pH = 14 - 4.92 = 9.08
What is the approximate pH of a 1.2 x 10^-5 M aqueous solution NaOH? A. 4.92 B. 7.50 C. 9.08 D. 12.45
pH = 3.82 pH = -log[H3O+] 3.82 = -log[H3O+] -3.82 = log[H3O+] Take antilog: 10^-3.82 = [H3O+] = 1.5 x 10^-4 [H3O+] = 1.5 x 10^-4 M
What is the concentration of H3O+ ions if a given solution has a pH of 3.82?
H2SO4 has 2 H+ groups, thus has the ability to produce 2 acid equivalents. So divide the molar mass by the number of equivalents: 98 g/mol/2 mol = 49 g Gram equivalent weight = 49 g
What is the gram equivalent weight of H2SO4? Molar mass H2SO4 = 98 g/mol.
Answer: B. Gram equivalent weight is the weight (in grams) that releases 1 acid or base equivalent from a compound. Because H3PO4 contains 3 protons, we find the gram equivalent weight by dividing the mass of one mole of the species by 3. The molar mass of phosphoric acid is 98 g/mol, so gram equivalent weight is 32.7 g.
What is the gram equivalent weight of phosphoric acid? A. 24.5 g B. 32.7 g C. 49.0 g D. 98.0 g
Answer: D. The question is asking for the pH, but because of the information given, we must first find the pOH and then subtract it from 14 to get the pH. Use the Henderson-Hasselbalch equation: pOH = pKb + log ([B-]/[BOH]) = 3.45 + log(70 mM/712 mM) = 3.45 - 1 = 2.45 If pOH = 2.45, the pH = 14 - 2.45 = 11.55
What is the pH of a solution with an ammonium concentration of 70 mM and an ammonia concentration of 712 mM? (Note: The pKb of ammonia is 3.45). A. 2.45 B. 4.45 C. 9.55 D. 11.55
[OH-] = 2.1 x 10^-3 M pOH = -log[OH-] pOH = -log[2.1 x 10^-3] pOH = 2.68 pH + pOH = 14 pH = 14 - pOH = 14 - 2.68 = 11.32 pH > 7 --> basic pH = 11.32 and the solution is basic
What is the pH of an aqueous NH3 solution with a concentration of OH- at 2.1 x 10^-3 M? Is the solution acidic, basic, or neutral?
pH = -log[H3O+] [H3O+] = 1.5 x 10^-4 pH = -log[1.5 x 10^-4] = 3.82 pH < 7 --> acidic pH of OJ is 3.82 and it is an acidic solution.
What is the pH of orange juice if it has a concentration of H3O+ ions at 1.5 x 10^-4? Is the solution classified as an acid, base, or neutral?
Kb of NH3 = 1.8 x 10^-5 pKb = -log[Kb] = -log[1.8 x 10^-5] = 4.74
What is the pKb of NH3? Kb = 1.8 x 10^-5
Sodium acetate (CH3COONa) pH > 7 basic A salt formed from a weak acid and a strong base will result in a basic solution.
What is the salt formed from an acid-base neutralization reaction between CH3COOH and NaOH? Would the pH of the solution be acidic, basic, or neutral? Explain why.
Ammonium chloride (NH4Cl) pH < 7 acidic A salt formed from a strong acid and a weak base will result in an acidic solution.
What is the salt formed from an acid-base neutralization reaction between HCl and NH3? Would the pH of the solution be acidic, basic, or neutral? Explain why.
Sodium Chloride (HCl) pH = 7 neutral A salt formed from a strong base and a strong acid will result in a neutral solution.
What is the salt formed from an acid-base neutralization reaction between HCl and NaOH? Would the pH of the solution be acidic, basic, or neutral? Explain why.
If a strong base is added, the acid in the buffer will neutralize the OH- ions. If a strong acid is added, the base in the buffer will neutralize the H3O+ ions.
What will happen if a strong base is added to a buffer solution? What about a strong acid?
Hydrolysis
When a salt ion is split into an acid and base by the presence of a water molecule, this reaction is referred to as a ______________ reaction.
100% ionization Right-side Products
When a strong acid/base undergoes equilibria, it undergoes _________________, which means that the total concentration of the acid turns into products. Because of this, equilibrium will lean to the __________-side of the reaction, favoring the _______________.
Ionization Protonated Left-side Reactants
When a weak acid undergoes equilibria, it does not undergo complete _______________. Weak acids are not able to donate their protons well, so it will stay mostly _______________. Because of this, equilibrium will lean towards the __________-side of the reaction, in favoring the ________________.
Bicarbonate ions (HCO3-) Carbonic acid (H2CO3) Acidic Carbonic acid (H2CO3) Bicarbonate ions (HCO3-) Basic
When an acidic substance enters our blood stream, the ________________ neutralizes the H3O+ ions, forming ________________ and water. This prevents the blood from becoming _________________. When a basic substance enters our blood stream, the ______________________ neutralizes the OH- ions, forming _________________ and water. This prevents the blood from becoming __________________.
Answer: B. Soluble hydroxides of Group IA and IIA metals are strong bases, eliminating (A.) and (D.). (B.) and (C.) are both weak bases; however, methylamine contains an alkyl group, which is electron-donating. This increases the electron density on the nitrogen in methylamine, making it a stronger (Lewis) base. Therefore, ammonia is the weakest base.
Which of the following bases is the weakest? A. KOH B. NH3 C. CH3NH2 D. Ca(OH2)
Answer: D. An amphoteric species is one that can act either as an acid or a base, depending on the environment. Proton transfers are classic oxidation-reduction reactions, so (A.) and (B.) are true. (C.) is true because many amphoteric species, such as water and bicarbonate, can either donate or accept a proton. (D.) is false, and thus the correct answer because amphoteric species can be either polar or nonpolar in nature.
Which of the following is NOT a characteristic of an amphoteric species? A. Amphoteric species can act as an acid or a base, depending on its environment B. Amphoteric species can act as an oxidizing or reducing agent, depending on its environment C. Amphoteric species are sometimes amphiprotic D. Amphoteric species are always nonpolar
Answer: B. First, convert the concentration to 5 x 10^-3 M. Next, because sulfuric acid is a strong acid, we can assume that, for the majority of sulfuric acid molecules (although not all), both portions will dissociate. The concentration of hydrogen ions is therefore 2 x 5 x 10^-3, or 10^-2. The equation for pH is pH = -log[H+]. If [H+] = 10^-2 M, then pH = 2.
Which of the following is closest to the pH of a solution containing 5 mM H2SO4? A. 1 B. 2 C. 3 D. 4
Answer: D. A Bronsted-Lowry base is defined as a proton acceptor. Ammonia, fluoride, and water - (A.), (B.), and (C.), respectively - each accepts a proton. (D.), HNO2, is a far better Bronsted-Lowry acid, donating a proton to solution.
Which of the following is not a Bronsted-Lowry base? A. NH3 B. F- C. H2O D. HNO2
Answer: A. Answering this question is simply a matter of knowing nomenclature. Acids ending in -ic are derivatives of anions ending with -ate, while acids ending in -ous are derivatives of anions ending in -ite. ClO3-, (B.), is chlorate because it has more oxygen than the other commonly occurring ion, ClO2-, which is named chlorite. Therefore, HClO3 is chloric acid. HClO4, (C.), represents chlorous acid. HClO, (D.), represents hypochlorous acid.
Which of the following represents chloric acid? A. HClO3 B. ClO3- C. HClO2 D. HClO