Chemistry II Unit 2

Aqueous ammonia can be used to neutralize sulfuric acid and nitric acid to produce two salts extensively used as fertilizers. They are...

(NH4)2SO4 and NH4NO3, respectively Aqueous ammonia is a weak base which reacts with acids to form salts. With sulfuric acid and nitric acid, it forms ammonium sulfate and ammonium nitrate, respectively, both of which are used as fertilizers.

The pH of 0.010 M aqueous aniline is 8.32. What is the percentage protonated?

0.021% pOH= 14-pH = 14-8.32= 5.68 10^-5.68/0.010 x 100= 0.021%

What is the [H+] concentration if a solution has pH= 8?

1 x 10^-8 [H+]= 10^8= 1 x 10^-8

What is [OH-] in a 0.0050 M HCl solution?

2.0 x 10^-12 M 1. Remember your equation for autoionization of water. Kw= [H+] x [OH-] = 1 x 10^-14 2. Plug in your concentration. [0.0050 m] x [OH-]= 1 x 10^-14 1 x 10^-14/0.0050= 2 x 10^-12

A 0.28 M solution of a weak acid is 3.5% ionized. What is the pH of the solution?

2.01 [H3O+]= [HA](%ionized) = 0.28×0.035=0.0098 M pH= -log(0.0098 M) = 2.01

Consider two solutions, one with pH= 7.8, the other pH= 9.3. How much more acidic is the first one than the second.?

31.6 times more acidic solution 1= pH= 1 x 10^-7.8 solution 2= pH= 1 x 10^-9.3 1 x 10^-7.8/1 x 10^-9.3=31.6

A 0.200 M solution of a weak monoprotic acid HA is found to have a pH of 3.00 at room temperature. What is the ionization constant of this acid?

5 x 10^-6 1. Use the generic equation. HA (aq) + H2O (l) ⇌ A- + H3O+ 2. Rice table. R: HA (aq) + H2O (l) ⇌ A- + H3O+ I: 0.200 - 0 0 C: -x - +x +x E: 0.200-x - x x [H+]= 1 x 10^-3 Ka= [A-][H3O+]/[HA] = (1 x 10^-3)^2/0.200 = 5 x 10^-6

A solution has [H+]= 1 x 10^-5 and [OH-] = 1 x10^-9 M. What is the pH and pOH? A. pH= 5, pOH= 9 B. pH= 0, pOH= 5 C. pH= 9, pOH= 5 D. pH= 5, pOH= 0

A. pH= 5, pOH= 9

Water can act as a A. acid B. base C. either an acid or a base D. Neither an acid or a base

C. either an acid or a base Amphoteric

HCl (aq) + H2O (l) → H3O+ (aq) + Cl- (aq) A weak acid reaction with water would be an equilibrium reaction. HF (aq) + H2O (l) ⇌ H3O+ (aq) + F- (aq) What is the equilibrium expression for HF in water?

K= [H3O+][F-]/[HF]

The Ka for chlorous acid, HClO2, is 1.2 x 10^-2. What is the equilibrium expression that accompanies this value when chlorous acid reacts with water?

Ka= [ClO2-][H3O+]/[HClO2]

What is the deprotonated form of nitrous acid?

NO2-

Which solution would have a higher concentration of [OH-]? (Hint: Write out your Kb expressions for these two.)

Propylamine

Identify the products of the following chemical reaction: Sr(OH)2 + 2HNO3 ⟶

Sr(NO3)2 + 2H2O acid + base → salt + water

When does the color change when an indicator is in a solution?

The color changes when the pH of the solution = the pKa of the indicator.

When you dissolve 1 mole of HBr in enough water to make 1 L solution, you expect the resulting solution to have _________. When you dissolve 1 mole of HNO2 in enough water to make 1 L solution, you expect the resulting solution to have _________-.

[H+] = 1 M [H+] < 1 M

Arrhenius Acid

a chemical compound that increases the concentration of hydrogen ions, H+, in aqueous solution

What is the pH of a solution made by mixing 0.050 mol of NaCN with enough water to make a liter of solution? Ka for HCN is 4.9 x 10^-10.

pH= 11 1. Remember the equation for Kw. Kw= Ka x Kb 4.9 x 10^-10 x Kb= 1 x 10^-14 Kb= 2.041 x 10^-5 2. Use Kb to solve for the concentration of OH in solution. [OH-]= sqrt Kb x Cb sqrt 2.041 x 10^-5 x 0.0050 = 1.01 x 10^-3 3. Remember that pH is going to be equal to 14- the -log[OH-]. pH= 14 - log[1.01x10^-3] = 11

Hydroxylamine is a weak molecular base with Kb = 6.6 x 10^-9. What is the pH of a 0.0500 M solution of hydroxylamine?

pH= 9.26 Hydroxylamine is a weak base, so we can use the equation below to approximate [OH-]. This approximation works because Kb is small and the concentration is not. [OH−]=√Kb⋅Cb=√(6.6×10^−9 x (0.0500)= 1.82×10^−5M [OH−]=Kb⋅Cb=(6.6×10^−9)(0.0500)=1.82×10^−5M pH=14−pOH=14−[−log(1.82×10−5)]=9.26

Strong bases

LiOH NaOH KOH RbOH CsOH Ca(OH)2 Sr(OH2) Ba(OH)2

A base is a _________ ________.

proton acceptor

For an anion to act as a base it must be

the conjugate base of a weak cid

An aqueous solution is prepared with 2 moles of HCl and 1 mole of Ca(OH)2. The resulting solution contains mainly...

water, Cl- ions, and Ca2+ ions. 1 mole of Ca(OH)2 reacts with 2 moles of HCl. So, there will be no Ca(OH)2 nor HCl left, only water and the salt CaCl2(aq) (which will exist as Ca2+ and Cl- ions). HCl is a strong acid and Ca(OH)2 is a strong base so they will completely dissociate in solution.

The stronger the acid, the ____________ its conjugate base.

weaker

Autoionization of water

when pure water reacts with itself to for hydronium and hydroxide ions 2 H2O (l) ⇌ H3O+ (aq) + OH- (aq)

pOH=

-log[OH-]

Solution A has a pH of 2. Solution B has pH of 4. What is true?

..

Please choose the products for the reaction in which HNO2 reacts with water.

H3O+ NO2-

percent ionization

[HA] ionized/[HA] initial x 100

[H+] > [OH-]

acidic pH < 7

A strong acid (or base) is one which...

dissociates completely in aqueous solution. Strong acids completely dissociates in aqueous solution.

Monoprotic

one acidic proton

pH=

pH= -log[H+]

[OH-]=

10^-pOH

The unionized form of an acid indicator is yellow and its anion is blue. The Ka of this indicator is 10^-5. What will be the color of the indicator in a solution of pH 3?

yellow The color change range is pH = pKa ± 1, or pH 4 - 6 in this case. At pH values above 6, the indicator will be ionized and at pH values below 4 the indicator will be unionized. Therefore, at pH 3, we expect the unionized, yellow version of the indicator.

[H+]=

10^-pH

What is the pH of a 0.25 M solution of Barium Hydroxide?

13.69

If the value of Kb for pyridine (C5H5N) is 1.8 x 10-^9, calculate the equilibrium constant for the following reaction: C5H5NH+(aq) + H2O(l) ⟶ C5H5N(aq) + H3O+(aq)

5.6 x 10^-6 Kw= Ka x Kb= 1 x 10^-14 Ka x 1.8 x 10^-9= 1 x 10^-14 1 x 10^-14/1.8 x 10^-9= 5.6 x 10^-6

Which solution has the highest pH? A. 0.1 M KClO, Ka for HClO is 3.5 x 10^-8 B. 0.1 M of KCl, Ka for HCl is VERY LARGE!! C. 0.1 M of KNO2, Ka for HNO2 is 4.5 x 10^-4 D. 0.1 M KCH3COO, Ka for CH3COOH is 1.8 x 10^-5

A. 0.1 M KClO, Ka for HClO is 3.5 x 10^-8 The solution with the highest pH would be the most basic. As all of the solutions have the same concentration and are conjugate bases of monoprotic acids, it all comes down to the Ka values of their conjugate acids. As we are looking for the most basic solution, we are looking for the strongest base. The strongest base would have the weakest conjugate acid. Therefore, we are looking for the smallest Ka, and that belongs to HClO. Therefore, a 0.1 M KClO solution would be the most basic and therefore have the largest pH.

The term "Ka for the ammonium ion" describes the equilibrium constant for which of the following reactions? A. NH4+ + H2O ⇌ NH3 + H3O+ B. NH3 + H2O ⇌ NH4+ + OH- C. NH4Cl(solid) + H2O ⇌ NH4+ + Cl- D. NH4+ + OH- ⇌ NH3 + H2O

A. NH4+ + H2O ⇌ NH3 + H3O+

If you dissolve HCl in water, which will be true? A. [H+]>[OH-] B. [H+]<[OH-] C. [H+] = [OH-]

A. [H+]>[OH-]

Which of the following is true in pure water at any temperature? A. [H3O+] = [OH-] B.Kw decreases with increasing temperature. C. [H3O+][OH-] = 1.0 x 10-14 D. pH = 7.0

A. [H3O+] = [OH-] Kw is shown to INCREASE with increasing temperature. pH = 7 is only true when water is at 24°C. [H3O+][OH-] = Kw, which increases with temperature. At high temperatures, pH can be less than 7. Thus [H3O+] = [OH-] is the only case that is true.

According to the Bronsted-Lowry concept of acids and bases, which of the following statements about a base is NOT true? A. base reacts with an acid to form a salt. B. A base must contain a hydroxide group. C. A base will share one of its electron pairs to bind H+. D. If a base is strong, then its conjugate acid will be relatively weaker.

B. A base must contain a hydroxide group. According to the Bronsted-Lowry definition, a base must be a proton acceptor. It does not need to contain a hydroxide group.

Which of the following substances is a strong acid? A. HF B. H2SO4 C. H3PO4 D. HSO3 E. H2CO3

B. H2SO4

What is the conjugate base of CH3CH2COOH? What is the conjugate acid of ClO-? A. CH3CH2CO+, HClO B. CH2CH2COOH, HClO C. CH3CH2COO-, HClO D. CH3CH2COOH2+. Cl-

C. CH3CH2COO-, HClO Conjugate base must have lost its proton, and conjugate acid must gain a proton

Which of these is an acid? A. CH3CH2OH B. CH3CH2COCH3 C. CH3CH2COOH D. CH3CH2COOCH3

C. CH3CH2COOH

Which one of the following combinations is NOT a buffer solution? A. HCN and NaCN B. CH3COOH and NaCH3COO C. HBr and KBr D. NH3 and (NH4)2SO4

C. HBr and KBr Strong acids dissociate 100%. Therefore, strong acids cannot exist in solution along with their conjugate bases and, by definition, cannot form buffer solutions. As HBr is a strong acid, it cannot form a buffer.

Solution A has a pH of 2. Solution B has a pH of 4. Which is true? A. [H+] in solution A is twice the [H+] in solution B. B. [H+] in solution B is twice the [H+] in solution A. C. [H+] in solution A is 100 times the [H+] in solution B. D. [H+] in solution B is 100 times the [H+] in solution A.

C. [H+] in solution A is 100 times the [H+] in solution B. Having a pH of 4 means that solution B is less acidic than solution A, meaning that the [H+] in solution A must be 100 times higher, since we use the log scale. A= [H+] = 1 X 10^-2 B= [H+] = 1 X 10^-4 1 X 10^-2/1X10^-4= 100

Identify the list in which all salts produce a basic aqueous solution. A. AlCl3, Zn(NO3)2, KClO4 B. AgNO3, NaCHO2, CrI3 C. NH4Cl, C6H4NH3NO3, FeI3 D. KCH3COO, NaCN, KF

D. KCH3COO, NaCN, KF Salts dissociate in water. While the cations in this case are spectator ions, the anions have the possibility to act as a base. For an anion to act as a base, it must be the conjugate base of a WEAK acid. For all salts in a list to produce a basic aqueous solution, the anions of each salt cannot be the conjugate base of a strong acid. For instance... NO3- is the conjugate base of HNO3 (nitric acid), a strong acid. Cl- is the conjugate base of HCl (hydrochloric acid), a strong acid. ClO4- is the conjugate base of HClO4 (perchloric acid), a strong acid.

Which of these is a base? A. Li2CO3 B. K2Cr2O7 C. NH4CH3CH2COO D. NH2CH2CH3

D. NH2CH2CH3

Which of the following substances is a weak acid? A. HBr B. HNO3 C. HClO3 D. H2SO4 E. HI F. H2CO3 G. HClO4 H. HCl

F. H2CO3

[H+] < [OH-]

basic solution pH>7

Lewis Acid

electron pair acceptor

Lewis Base

electron pair donor

Polyprotic

more than one acidic proton

Strong acid + strong base

neutral salt

[H+] = [OH-]

neutral solution pH = 7

Identify the salt that is produced from the acid-base neutralization reaction between potassium hydroxide and acetic acid.

potassium acetate CH3COOH + KOH → KCH3COO + H2O KCH3COO= potassium acetate

Blood contains a buffer of carbonic acid (H2CO3) and hydrogen carbonate ion (HCO3-) that keeps the pH at a relatively stable 7.40. What is the ratio of [HCO3-] / [H2CO3] in blood? Ka1 = 4.30x10^-7 for H2CO3. (Hint: Assume [CO3^2-] = 0)

10.8 1. Write out the equation. H2CO3 + H2O ⇌ HCO3- + H3O+ 2. Given pH, calculate your concentration of H+. pH= -log[H+]= 7.4 [H+]= 3.98 x 10^-8 3. Use your [H+] and Ka to determine your ratio. Ka= [HCO3-][H+ or 3.98 x 10^-8]/[H2CO3]= 4.3 x 10^-7 [HCO3-]/[H2CO3]= 4.3 x 10^-7/3.98 x 10^-8= 10.8

Strong acids

HCl (hydrochloric acid) HBr (hydrobromic acid) HI (hydroiotic acid) HClO4 (perchloric acid) HClO3 (chloric acid) H2SO4 (sulfuric acid) HNO3 (nitric acid)

The conjugate base of H2SO4 is:

HSO4- H2SO4 only has one acidic proton, so it's conjugate base only loses one hydrogen.

Spectator ions

Ions that do not take part in a chemical reaction and are found in solution both before and after the reaction

The Kb for aniline, C6H5NH2, is 4.3 × 10^-10 . What is the equilibrium expression that accompanies this value?

Kb= [C6H5NH3+][OH-]/[C6H5NH2]

What is the percent ionization for a weak acid HX that is 0.40 M? Ka = 4.0 x 10^-7.

0.10% 1. RICE Table R: HX + H2O ⇌ A- + H3O+ I: 0.40 - 0 0 C: -x - +x +X E: 0.40-x - x x 2. Products over reactants. (x^2)/(0.40-x) x= 4 x 10^-4 4 x 10^-4/0.40 x 100= 0.10%

Calculate the percent ionization of 0.3 moles HNO2 dissolved in 1 L of a 0.4 M NO2- solution. Assume no volume change due to the addition of HNO2. The Ka of HNO2 is 4.5x10^-4. What is the pH of this solution?

0.11% pH= 3.47 3.5% 1. Write out the reaction. HNO2 (aq) + H2O (l) ⇌ NO2 (aq) + H3O (aq) 2. Make an Ice table. HNO2 (aq) + H2O (l) ⇌ NO2 (aq) + H3O (aq) I 0.3 - 0.4 0 C -X +X +X E 0.3-x 0.4 +X X 3. Write out your equilibrium equation. [NO2][H30]/[HNO2] = 4.5 x 10^-4 0.4x/0.3= 4.5 x 10^-4 x= 3.37 x 10^-4= [H+] *Note: you can ignore x's that are added because they will be relatively small in comparison to the amount of moles of reactant you have. 4. Calculate percent ionization. [HA] ionized/[HA] initial x 100 3.37 x 10^-4/0.3 x 100= 0.11% 5. Calculate pH. -log[3.37 x 10^-4] = 3.47

A solution is 0.30 M in NH3. What concentration of NH4Cl would be required to achieve a buffer solution with a final pH of 9.0? Kb = 1.8x10^-5 for NH3.

0.54 M 𝐾𝑤=𝐾𝑎⋅𝐾𝑏=𝐾𝑎(1.8×10^−5)= 1 x 10^−14 𝐾𝑎=5.56×10^−10 9= -log(5.56 x 10^-10) + log (0.30/acid) [NH4Cl] = 0.30 M

Which of these would be viable buffer solutions? A. 1 M HF and 0.0001 M F- B. 1 M HCl and 1 M Cl- C. 1 M NH3 and 1.5 M NH4+ D. 1 M HF and 1 M F-

1 M NH3 and 1.5 M NH4+ 1 M HF and 1 M F-

What would be the final pH if 0.0100 moles of solid NaOH were added to 100mL of a buffer solution containing 0.600 molar formic acid (ionization constant = 1.8x10^-4) and 0.300 M sodium formate?

3.65 0.0100 moles of NaOH in 100 mL of solution gives 0.1 M NaOH. This will neutralize with the 0.600 M formic acid to reduce the amount of acid and increase the amount of sodium formate. 1. Calculate molarity. 0.0100/0.1 L= 0.1 M NaOH 2. Since your base will neutralize your acid, subtract the concentration of NaOH from the concentration of acid and add it to your salt. [formic acid] = 0.600 - 0.100 = 0.500 M [sodium formate] = 0.300 + 0.100 = 0.400 M 3. The ionization constant provided is for formic acid, so it is a Ka. Use the Henderson-Hasselbech equation to solve for pH. pH = pKa + log[Base/acid] pH= -log(1.8 x 10^-4) + log(0.400/0.500)= 3.65

Identify the products of the following chemical reaction: 3LiOH + H3PO4 ⟶

Li3PO4 + 3H2O acid + strong base → salt + water

Please choose the products for the reaction in which CH3NH2 reacts with water: CH3NH2 (aq) + H2O (l) ⇌ ?

OH- CH3NH2+

Ternary acid

contains 3 elements

Carboxylic acid

contains a carboxyl group

An acid is a ________ __________.

proton donor

3 ways to make a buffer

1. Mix a weak acid and conjugate base, or weak base with conjugate acid. 2. Mix a strong base and excess weak acid. When you do this, part of the weak acid is neutralized and converted to its conjugate base. The remaining part of the weak acid stays as the weak acid. 3. Mix a strong acid with excess weak base. When you do this, part of the weak base is neutralized and converted to its conjugate acid. The remaining part of the weak base stays as the weak base.

How many moles of Ca(OH)2 are needed to neutralize three moles of HCl?

1.5 moles 1. Write out the equation. For an acid-base neutralization, we need one mole of H+ ions for every mole of OH- ions. Ca(OH)2 + 2HCl → CaCl2 + 2H2O 2. Use the coefficients to establish ratio needed. 1 mole of Ca(OH2) and 2 moles of HCl 3 mol HCl x 1 mol Ca(OH)2/2 moles of HCl = 1.5 mol Ca(OH)2

What is the pH of a 0.23 M solution of potassium generate (KR-COO)? Ka for the generic acid R-COOH is 2.7 x 10^-8.

10.47 1. In essence, "potassium generate" is a weak generic base of the form (KA). This means we need Kb for the weak base A-. Kw= Ka⋅Kb=(2.7×10^−8)x Kb = 1 x 10^−14 Kb= 3.7037 × 10^−7 2. Now we can use the approximation equation to find [OH-]. [OH−]= √Kb⋅Cb= √(3.7037×10^−7) [OH−]= √Kb⋅Cb= √(3.7037×10−7) √(3.7037×10−7) x 0.23 = 0.000291865 M = [OH-] 14- -log[0.000291865]=10.47

What is the pH of a solution which is 0.600 M in dimethylamine ((CH3)2NH) and 0.400 M in dimethylamine hydrochloride ((CH3)2NH2Cl)? Kb for dimethylamine = 7.4x10^-4.

11.05 𝐾𝑤=𝐾𝑎x 𝐾𝑏 =𝐾𝑎 x (7.4×10−4)= 1 x 10^−14 𝐾𝑎=1.35×10−11 𝑝𝐻=𝑝𝐾𝑎+log([𝑏𝑎𝑠𝑒][𝑎𝑐𝑖𝑑]) 𝑝𝐻=−log(1.35×10−11) + log(0.6000/0.400)=11.05

What is the pH of a 0.1 M Ba(OH)2 aqueous solution?

13.3 1. Notice that Ba(OH)2 is a strong base and that the concentration of OH will be 0.1 x 2. 0.1 x 2= 0.2 2. Calculate pOH. -log(0.2)= 0.69 3. Realize that you've calculate pOH, so you'll need to subtract from pH. 14- 0.7= 13.3

What is the pH of a 0.25 M solution of potassium hydroxide?

13.4 1. Recognize that KOH is a strong base. This means; [KOH]o=[OH-]eq = 0.25 M 2. Use your Kw equation to solve for your concentration of H3O+. Kw= [H3O+][OH-] = 1 x 10^-14 [H3O+][0.25 M]= 1 x 10^-14 [H3O+]= 4 x 10^-14 2. Calculate pH by using the -log equation. pH= -log(4 x 10^-14)= 13.4

What is [H3O+] when [OH-] = 3.3 x 10^-9 M? A. 3.0 x 10^-6 M B. 3.3 x 10^-9 M C. 1.0 x 10^-7 M D. 3.3 x 10^-5 M

3.0 x 10^-6 M 1. Remember the equation for autoionization of water. Kw= [H3O+][OH-]= 1 x 10^-14 2. Plug in your unit. [H3O+][3.3 x 10^-9]= 1 x 10^-14 3. Divide by constant. 1 x 10^-14/3.3 x 10^-9= 3.0 x 10^-6

Calculate the percent ionization when you dissolve 0.5 moles HF in enough water to make 1 L of solution. Ka = 6.3 x 10^-4 Assume negligible volume change. (Hint: Use your ICE table and equilibrium constant equation to figure out how much HF ionized.)

3.5% 1. Write out the reaction. HF (aq) + H2O (l) ⇌ H3O (aq) + F (aq) 2. Make an Ice table. HF (aq) + H2O (l) ⇌ H3O (aq) + F (aq) I 0.5 - 0 0 C -X +X +X E 0.5-x X X 3. Write out your equilibrium equation. [H3O+][F-]/[HF] = 6.3 x 10^-4 X^2/0.5-x=6.3 x 10^-4 *Note: you can ignore the x on the bottom because it will be relatively small in comparison to the amount of moles of reactant you have. 4. Cross multiply to solve for x. X^2/0.5-x=6.3 x 10^-4 X^2= 3.15 x 10^-4 sqrt 3.15x10^-4= 0.0177 = x 5. Calculate percent ionization. [HA] ionized/[HA] initial x 100 0.0177/0.5 x 100= 3.5%

A buffer solution is made by dissolving 0.45 moles of a weak acid (HA) and 0.33 moles of KOH into 710 mL of solution. What is the pH of this buffer? Ka = 6x10^-6 for HA.

5.66 Divide the number of moles of the weak acid and strong base by 0.71 L to find the molarity of each. This gives a solution that is 0.6338 M in HA and 0.4648 M in KOH. You must subtract the 0.4648 M KOH from the 0.6338 M HA because the strong base will neutralize the weak acid. You therefore would make a solution that is 0.4648 M in A- and 0.1690 M HA. pH= pKa + -log [A-]/[HA] pH= -log[6 x 10^-6] + -log[0.4648/0.1690] = 5.66

What is the pH of a solution that contains 11.7g of NaCl for every 200 mL of solution?

7 NaCl completely dissociates in water to give Na+ and Cl-, neither of which hydrolyzes. So, in aqueous NaCl, the H3O+ and OH- ions result completely form the autoionization of water.

A buffer was prepared by mixing 0.200 moles of ammonia (Kb = 1.8x10^-5) and 0.200 moles of ammonium chloride to form an aqueous solution with a total volume of 500 mL. 250 mL of the buffer was added to 50.0 mL of 1.00 M HCl. What is the pH of this second solution?

8.78 50.0 mL of a 1.00 M HCl solution contains 0.0500 moles of HCl acid. Since only half of the buffer solution is added to the final solution (250 mL of the 500 mL), then only 0.100 moles of ammonia and ammonium chloride are actually added to the final solution. The HCl will neutralize the ammonia to form ammonium chloride. 0.100 - 0.0500 = 0.050 moles ammonia 0.100 + 0.0500 = 0.150 moles ammonium chloride The final solution volume comes from 250 mL of the buffer and 50.0 mL of the HCl acid, giving a total volume of 300 mL. Dividing the moles above by the volume gives 0.1667 M ammonia and 0.5 M ammonium chloride. pOH= pKb + log [Base/acid] pOH= -log(1.8 x 10^-5) + log (0.5/0.1667)= 5.22

What is the pH in a solution made by dissolving 0.100 moles of sodium acetate (NaCH3COO) in enough water to make one liter of solution? Ka for CH3COOH is 1.80 x 10^-5.

8.87 Sodium acetate is a salt that completely dissociates in solution giving acetate ion (CH3COO-) as a base. 1. Recognize that you're given Ka in the problem, so you will need to solve for Kb. Kb= Kw/Ka Kb= 1 x 10^-14/1.8x10^-5.55 x 10^-20 2. Use your Kb and concentration of salt to calculate your OH. [OH-] = sqrt Kb x Cb sqrt 5.55 x 10^-20 x 0.100 = 7.454 x 10^-6 M = [OH] 3. Calculate pH by subtracting the -log[OH] from 14. pH= 14-[-log(7.454 x 10^-6)]= 8.87

Which of these solutions would have the highest buffer capacity? A. solution that is 1 M HF and 1 M F- B. A solution that is 0.1 M HF and 0.1 M F- C. A solution that is 0.01 M HF and 0.01 M F-

A. solution that is 1 M HF and 1 M F-

A water solution of sodium acetate is basic because... A. the acetate ion acts as a Bronsted-Lowry base in a reaction with water. B. sodium acetate is only weakly ionized. C. the conjugate base of the acetate ion is a strong base. D. The statement is false. A water solution of sodium acetate is acidic.

A. the acetate ion acts as a Bronsted-Lowry base in a reaction with water. In water, sodium acetate dissociates into sodium ions (spectator ions) and acetate ions which hydrolyze to produce hydroxide ions. CH3COO- + H2O ⇌ CH3COOH + OH- The acetate ion accepts a proton from a water molecule, and is thus a Bronsted-Lowry base.

Assume that five weak acids, identified only by numbers (1, 2, 3, 4, and 5) have the following ionization constants: A1 - Ka = 1.0 x 10^-3 A2 - Ka = 3.0 x 10^-5 A3 - Ka = 2.6 x 10^-7 A4 - Ka = 4.0 x 10^-9 A5 - Ka = 7.3 x 10^-11 The anion of which acid is the strongest base?

A4 - Ka = 4.0 x 10^-9 The smallest Ka= the highest Kb The strength of an acid and its conjugate base are inversely related. So, the weaker the acid, the stronger its conjugate base. To find the strongest conjugate base, we are looking for the weakest acid, or the acid with the smallest Ka value. That acid is acid number 5.

Which of these reactions would result in a buffer solution? A. Reacting 100 mL of 1 M of NaOH with 200 mL of 1 M HCl. B. Reacting 100 mL 1 M of HCl with 200 mL of CH3NH2. C. Reacting 100 mL of 1 M NaOH with 200 mL of 1 M HF. D. Reacting 100 mL of 1 M HCl with 200 mL of 1 M HF.

B. Reacting 100 mL 1 M of HCl with 200 mL of CH3NH2. C. Reacting 100 mL of 1 M NaOH with 200 mL of 1 M HF.

Which is NOT a conjugate acid-base pair, respectively? A. HCN : CN- B. SO4 2- : HSO4- C. H2O : OH- D. H3O+ : H2O

B. SO4 2- : HSO4- In the Bronsted-Lowry sense, a base is a proton acceptor and an acid is a proton donor. HSO4- is a proton donor and an acid. Therefore, this is a conjugate base-acid pair, not a conjugate acid-base pair.

Which pH represents a solution with 1000 times higher [OH-] than a solution with a pH of 5? A. pH = 6 B. pH = 8 C. pH = 4 D. pH = 7

B. pH = 8 1. Given [OH-], calculate pH. [H+]-[OH-]= pH 14-5= 9 2. We're looking for a solution with 1000 times the hydroxide concentration of the solution above. 1000 x 10-^9 = 10^-6 M OH-. pOH=−log[OH−]=−log(10^−6)=6 14-6=8

HCN is classified as a weak acid in water. This means that it produces... A. 100% of the maximum number of possible hydronium ions. B. no hydronium ions. C. a relatively small fraction of the maximum number of possible hydronium ions. D. a relatively large fraction of the maximum number of possible hydronium ions.

C. a relatively small fraction of the maximum number of possible hydronium ions. In water, only a small percentage of a weak acid hydrolyzes, producing H3O+ ions.

Which of the following mixtures will be a buffer when dissolved in a liter of water? A. 0.2 mol HBr and 0.1 mol NaOH B. 0.1 mol Ca(OH)2 and 0.3 mol HI C. 0.3 mol NaCl and 0.3 mol HCl D. 0.2 mol HF and 0.1 mol NaOH

D. 0.2 mol HF and 0.1 mol NaOH The answer choices with Ca(OH)2 and HBr contain mixtures of strong acids and strong bases. Mixing strong with strong cannot form a buffer solution. At least one of the compounds has to be a weak acid or weak base. The choice with NaCl seems to mix an acid with its conjugate base, but HCl is a strong acid. Remember strong acids mixed with their conjugate bases cannot be buffer solutions because the acid dissociates 100% and, therefore, cannot exist in solution along with its conjugate. The answer choice with HF and NaOH is the only choice that would form a buffer. The NaOH will neutralize SOME of the HF, but as there is more HF than NaOH, it cannot neutralize all of it. Therefore, HF (a weak acid) will exist in solution with its conjugate base F- (a weak base). This is the definition of a buffer.

In the forward reaction HCN + H2O ⇌ CN- + H3O+, the Bronsted-Lowry acid is...

HCN In the Bronsted Lowry definition, acids are proton donors. In this reaction, HCN, a weak acid, donates it proton to water to form a hydronium ion.

What is the conjugate acid of NO3-?

HNO3

When an acid and base neutralize each other, the products are generally water and...

a salt The general format for neutralization reactions is: strong acid + strong base → salt + water

Arrhenius Base

a substance that increases the concentration of hydroxide ions in aqueous solution

The larger the Ka of the acid, the __________ the Kb of its conjugate base.

smaller