Aqueous Equilibria

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 sulfu- ric acid and nitric acid, it forms ammonium sulfate and ammonium nitrate, respectively, both of which are used as fertilizers.

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

H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a solution that is 0.100 M H2SO4? Account for the ionization of both protons.

0.963

Determine the molar solubility of some salt with the generic formula AB2 if Ksp = 2.56x10^2

4 M

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

1.5 nHCl = 3 mol For acid base neutralization we need one mole of H+ for every mole of OH−. Therefore the balanced equation is 2 HCl + Ca(OH)2 → CaCl2 + 2 H2O X mol Ca(OH)2 = 3 mol HCl × 1 mol Ca(OH)2/ 2 mol HCl = 1.5 mol Ca(OH)2

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)

10.75 Given data --> pH = 7.40, Ka = 4.30 * 10-7 The general representation of equilibrium constant is given by, Ka = [H+][A-] / [HA] for blood the above expression deduces to, Ka = [H+][HCO3-] / [H2CO3] -------- (1) on rearranging equation (1) for the ratio is given by, [HCO3-] / [H2CO3] = Ka / [H+] ---------(2) by substituting the given values in equation (2) we get, {pH = -log [H+] = 7.40 so [H+] = 10-7.4 = 4*10-8 M } [HCO3-] / [H2CO3] = (4.30 *10-7) / (4*10-8) [HCO3-] / [H2CO3] = 10.75 = 11 (approx)

50.0 mL of 0.0018 M aniline (a weak base) is titrated with 0.0048 M HNO3. How many mL of the acid are required to reach the equivalence point?

18.8 mL

What would be the molar solubility of Li3PO4 (Ksp = 2.37 x 10^-4) in a 1M LiCl solution?

2.37 x 10^-4

A hypothetical compound MX3 has a molar solubility of 0.00562 M. What is the value of Ksp for MX3?

2.69 x 10^-8

What is the pH at the half-stoichiometric point for the titration of 0.22 M HNO2(aq) with 0.1 M KOH(aq)? For HNO2, Ka = 4.3x10-4.

3.37

One liter of a buffer composed of 1.2 M HNO2 and 0.8 M NaNO2 is mixed with 400 mL of 0.5 M NaOH. What is the new pH? Assume the pKa of HNO2 is 3.4.

3.4 First, calculate how many moles of added base are introduced to the system: (0.4 L)(0.5 M) = 0.2 moles of NaOH The strong base will interact with the HNO2 to create more NO2^- (this is a very important concept!). The NaOH transforms 0.2 moles of HNO2 into NO2-. Now our new HH equation looks like this: pH = 3.4 + log(1.0/1.0) The log of 1 is just zero so the pH = pKa. Some students are worried about the volume change (good thought!), however, in this case, the equation divides the concentration of A- by the concentration of HA. In this problem, they are in the same solution and so the volumes associated with both of these concentrations are identical and dividing the concentration values cancels out the identical volume values.

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

What is the net ionic equation for the reaction between aqueous solutions of Na3PO4 and CuSO4?

3Cu2+ + 2PO43 ⟶ Cu3(PO4)2

The acid form of an indicator is yellow and its anion is blue. The Ka of this indicator is 10^-5. What will be the approximate pH range over which this indicator changes color?

4 < pH < 6 The pKa of this indicator is 5, so the indicator will change colors around pH 5. Thus you would expect a color change between pH 4 and pH 6.

A solution of AgI contains 1.9 M Ag+. Ksp of AgI is 8.3 x 10^-17. What is the maximum I- concentration that can exist in this solution?

4.4x10^-17 M

What is the pH of a 3.4x10^-8 M HClO4 solution at room temperature?

6.9

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 **********

What is the pH at the equivalence point in the titration of 10.0 mL of 0.35 M unknown acid HZ with 0.200 M NaOH? Ka = 2.4x10-7 for the unknown acid HZ

9.86 How many moles of HZ are present? How many moles of NaOH are required to reach the equivalence point and what volume of base would you have to add? HZ will be neutralized to its conjugate base. What is the concentration of the conjugate base? How will this conjugate base react in a neutral solution? (Note: You'll need to know Kb for the conjugate base.)

CaSO4 has a Ksp = 3x10^-5. In which of the following would CaSO4 be the most soluble? A. pure water B. 0.5 M K2SO4(aq) C. CaSO4 would have the same solubility in all three of these solutions D.1 .0 M CaCl2(aq)

A. pure water

Rank the following salts from least to most molar solubility: BiI Ksp = 7.7x10^-19 Cd3(AsO4)2 Ksp = 2.2x10^-33 AlPO4 Ksp = 9.8x10^-21 CaSO4 Ksp = 4.9x10^-5

AlPO4 < BiI < Cd3(AsO4)2 < CaSO4

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

B. HBr and KBr A buffer must contain a weak acid/base conjugate pair. HBr/Br− is a strong acid conjugate pair. CH3COOH/CH3COO−, HCN/CN− and NH+4 /NH3 are weak acid/base conjugate pairs.

Determine if a precipitate will form when 0.96g Na2CO3 is combined with 0.2g BaBr2 in a 10L solution. (For BaCO3, Ksp = 2.8x10^-9).

BaCO3 precipitates

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

D. 0.2 mol HF and 0.1 mol NaOH Eliminate answers that are obviously incor- rect. The choice with "0.2 mol HBr" and "0.1 mol Ca(OH)2" are strong acids and strong bases respectively; therefore, NOT buffers. The choice with "0.3 mol NaCl" is a combina- tion of spectator ions and a strong acid; this does not form a buffer. Remaining for cal- culation are choices with "0.4 mol NH3" and "0.2 mol HF". Now perform the neutralizaton calculations on the remaining possibilities: Choice with 0.4 mol NH3 ******* look up again

Pure water is saturated with PbCl2. In this saturated solution, which of the following is true? A. Ksp = [Pb2+][Cl-] B. [Pb2+] = 0.5[Cl-] C. [Pb2+] = [Cl-] D. Ksp = [Pb2+]^2[Cl-]

D. Ksp = [Pb2+]^2[Cl-]

What ions are present in solution after aqueous solutions of Cu(NO3)2 and K2S are mixed? Assume we mixed stoichiometric equivalent amounts of both reactants and 100% reaction.

K+, NO3-

The Ksp equation for sodium bicarbonate (NaHCO3) should be written as:

Ksp = [Na+][H+][CO3^2-]

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

LI3PO4 + 3H2O Acid + Base → Salt + Water

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

Sr(NO3)2 + 2 H2O Sr(OH)2 is a base and HNO3 an acid; they create a salt and water.

A buffer is composed of NH3 and NH4Cl. How would this buffer solution control the pH of a solution when a small amount of a strong base is added?

The OH- reacts with the NH4+. The conjugate acid component (NH4+) of the buffer would react with the added base (OH-).

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

a salt The general format for neutralization reac- tions is acid + base → salt + water.

The buffer ____________ of a buffer solution is the total amount of acid or base that can be added to the buffer before ________ of ________ of the forms of the compound.

capacity neutralizing one If you add some acid or base to a buffer solution one of the forms of the buffer compound will react - not both!

When we titrate a weak base with a strong acid, the pH at the equivalence point will be...

pH<7 At the equivalence point, all of the strong acid will be consumed, leaving only its very weak (to the point of being neutral) conjugate base behind. However, the weak base will be converted into its corresponding conjugate acid, which will not be neutral. The resulting solution will then be acidic, having a pH < 7.

The Henderson-Hasselbalch formula allows us to determine if the majority of a compound will be in the protonated or deprotonated form at a given pH by comparing the pH of the solution to the _____. The exact _____ of protonated form and deprotonated form can be obtained from the Henderson-Hasselbalch equation.

pKa, Ratio The HH equation compares the pH of a solution to the pKa of the compound and allows us to calculate the exact ratio of the conjugate acid and base.

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? Recall the HH equation can be written using pH and pKa OR pOH and pKb. The Kb of NH3 is 1.8 x 10-5

0.54 M First, because a Kb value is given it might be easier to choose to use the basic form of the HH equation. Here pOH = pKb + log([HB+]/[B]). If the pH is 9.0 then the pOH is 5.0. Find the pKb by taking the negative log of the Kb value pKb = -log(1.8x10-5) = 4.74. So you have: pOH = pKb + log([HB+]/[B]) 5.0 = 4.74 + log(x/0.3) 0.26 = log(x/0.3) 100.26 = x/0.3 1.8 = x/0.3 0.54 = x

Identify which of the following statements that are true of buffer solutions: 1) A buffer solution could consist of equal concentrations of ammonia and ammonium bromide. 2) A buffer solution could consist of equal concentrations of perchloric acid (HClO4) and sodium perchlorate. 3) A buffer solution will change only slightly in pH upon addition of small amounts of acid or base. 4) In a buffer solution containing benzoic acid (C6H5COOH) and sodium benzoate (NaC6H5COO) the species that reacts with added hydroxide ion is the benzoate ion.

1, 3 Buffer solutions are often made of weak conjugate acid base pairs like ammonia and ammonium bromide (statement 1). Buffer solutions are useful because their pH changes very little with small additions of acid/base (statement 3). Buffer solutions are not made with strong acid base conjugate pairs (statement 2). In buffer solutions, the acid component reacts with addition of any base while the base component reacts with addition of any acid. In statement 4 the benozic acid would react with hydroxide not the benzoate anion. NH^+4 is the conjugate acid of the weak base NH3; C6H5COO- is the conjugate base of theweak acid C6H5COOH. HClO4 is a strong acid. OH -ions introduced into an acid buffer system removes the H3O+. Buffers are composed of comparative amounts of aweak acid/base and its conjugate and allows only a small change in pH when an acid or base is added.

A hypothetical ionic substance T3U2 ionizes to form T^2+ and U^3- ions. The solubility of T3U2 is 4.04x10-20 mol/L. What is the value of the solubility-product constant?

1.08x10^-97

For the titration of 50.0 mL of 0.020 M aqueous salicylic acid with 0.020 M KOH (aq), calculate the pH after the addition of 55.0 mL of the base. For salycylic acid, pKa = 2.97.

10.98

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 Ka, (CH3)2NH+2 = Kb, (CH3)2NH Applying the Henderson-Hasselbalch equation, pH = pKa + log [(CH3)2NH] [(CH3)2NH+2 ] =−logKw+log [(CH3)2NH] Ka [(CH3)2NH+2 ] = − log 1 × 10−14 + log 0.6 0.00074 0.4 = 11.0453 *********

The value of Ksp for SrSO4 is 2.8x10^-7. What is the solubility of SrSO4 in moles per liter?

5.3 x 10^-4

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 nHA = 0.45 mol Ka =6×10−6 for HA nKOH = 0.23 mol You must substract the 0.33 moles of KOH from the 0.45 moles of HA because the strong base will neutralize the weak acid. You therefore would make 0.23 moles of A− and be left with 0.22 moles of HA. You can now plug this ratio into the equilibrium equation or in the Henderson-Hasselbalch equation to get pH.

Consider the titration of 50.0 mL of 0.0200 M HClO(aq) with 0.100 M NaOH(aq). What is the formula of the main species in the solution after the addition of 10.0 mL of base?

ClO-

What is the pH of an aqueous solution that is 0.018 M C6H5NH2 (Kb = 4.3x10-10) and 0.12 M C6H5NH3Cl?

Use the Henderson Hasselbach equation: This equation can be written 2 ways: pH = pKa + log(base/acid) pOH = pKb + log(base/acid) Since you are given the Kb, let's use the second equation and find the pOH. pOH = pKb + log(base/acid) pOH = -log(4.3 x 10^-10) + log(0.12/.0.018) pOH = 10.19 Now: pH + pOH = 14 pH + 10.19 = 14 pH = 3.81

Aspartic acid is a polypeptide side chain found in proteins. The pKa of aspartic acid is 3.86. If this polypeptide were in an aqueous solution with a pH of 7, the side chain would have what charge?

negative Since the pH is greater than the pKa, the acid group will be deprotonated leaving the sidechain with a charge of minus 1.

Identify the salt that is produced from the acid-base neutralization reaction between potassium hydroxide and acetic acid (CH3COOH).

potassium acetate CH3CO2H(aq) + K+(aq) + OH−(aq) → K+(aq) + CH3CO−2 (aq) + H2O(l) Potassium acetate (CH3CO2K) is the salt.

Molar solubility is...

the number of moles that dissolve to give one liter of saturated solution.

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. 2 HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2 H2O(aq) 1 mole of Ca(OH)2 reacts with 2 moles of HCl, so there will be no Ca(OH)2 nor HCl left. The CaCl2(aq) will exist as Ca2+(aq) and Cl−(aq). The H+ from the HCl and the OH− from the Ca(OH)2 have all reacted. Only a miniscule amount of H+ and OH− remain from the autoionization of the water.

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