1. Which of the following mixtures are buffers and why? If it is a buffer, write
ID: 820486 • Letter: 1
Question
1. Which of the following mixtures are buffers and why?
If it is a buffer, write an equilibrium equation for the conjugate acid/base pair.
a. KF / HF
b. NH3 / NH4Br
c. KNO3 / HNO3
d. Na2CO3 / NaHCO3
2. A 200 mL sample of 0.1 M NH4Cl solution is mixed with 160 mL of 0.2 M ammonia, NH3, solution to make a buffer solution, where Kb = 1.79 x 10^-5 for NH3
a. What is the PKb for ammonia NH3
b. What is the pH of the resulting buffer solution?
c. What is the pH if 0.5 grams of NaOH is added to the buffer solution?
d. What is the pH if 0.5 grams of H2SO4 is added to the buffer.
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I am offering a lot of points for this, so please give complete answers and provide some work too? Thanks in advance ^_^
Explanation / Answer
1) Buffers consist of weak acids and their conjugate bases, thus the answer is:
a. KF / HF : HF(aq) <=> H+(aq) + F-(aq), HF is a weak acid and F- is the conjugate base
b. NH3 / NH4Br : NH4+(aq) <=> H+(aq) + NH3(aq), NH4+ is a weak acid and NH3 is the conjugate base
d. Na2CO3 / NaHCO3 : HCO3-(aq) <=> H+(aq) + (CO3)2-(aq), HCO3- is a weak acid and (CO3)2- is the conjugate base
(Note: KNO3/HNO3 is not a buffer as HNO3 is a strong acid)
(2) (a) pKb = -log Kb = -log(1.79 x 10^(-5)) = 4.747 = 4.75
(b) Initial moles of NH4+ = 200/1000 x 0.1 = 0.02 mol
Initial moles of NH3 = 160/1000 x 0.2 = 0.032 mol
pKa = 14 - pKb = 14 - 4.75 = 9.253
pH = pKa + log([NH3]/[NH4+])
= pKa + log(moles of NH3/moles of NH4+)
= 9.253 + log(0.032/0.02) = 9.46
(c) Moles of NaOH added = mass/molar mass of NaOH = 0.5/40.00 = 0.0125 mol
NH4+ + NaOH => Na+ + NH3 + H2O
Moles of NH4+ = 0.02 - 0.0125 = 0.0075 mol
Moles of NH3 = 0.032 + 0.0125 = 0.0445 mol
pH = pKa + log([NH3]/[NH4+])
= pKa + log(moles of NH3/moles of NH4+)
= 9.253 + log(0.0445/0.0075) = 10.03
(d) Moles of H2SO4 added = mass/molar mass of H2SO4 = 0.5/98.08 = 0.005098 mol
2 NH3 + H2SO4 => 2 NH4+ + (SO4)2-
Moles of NH4+ = 0.02 + 2 x 0.005098 = 0.030196 mol
Moles of NH3 = 0.032 - 2 x 0.005098 = 0.021804 mol
pH = pKa + log([NH3]/[NH4+])
= pKa + log(moles of NH3/moles of NH4+)
= 9.253 + log(0.021804/0.030196) = 9.11
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