Question

A solution has 0.100 M HC7H5O2 and 0.200 M Ca(C7H5O2)2. The solution volume is 5.00 L. What is the pH of the solution after 2.00 g NaOH is added? The Ka for HC7H5O2 is 6.3 × 10-5.

If 0.02 moles of HCl was added to the solution described in Question 2, what change do you expect to occur in the pH?

Calculate the milligrams of Ag2CO3 in 250 mL of a saturated solution of Ag2CO3. The Ksp of Ag2CO3 is 8.1 * 10-12.

Answer 1

Q1

mol of NaOH =mass/MW = 2/40 = 0.05 mol

mol of acid initilly = MV = 0.1*5 = 5

mol of A-2 = 2*MV = 2*0.2*5 = 2

after addition of 0.05 mol of base

mol of acid left = 5-0.05 = 4.95

mol of conjguate formed = 2 + 0.05 = 2.05

pH = pKa + log(a-/HA)

ka = 6.3*10^-5 --> to pKa via -log(Ka)

pKa = 4.20 for benzoic acid

pH = 4.20 + log(2.05/4.95) = 3.82

if we added 0.02 mol of HCl ...

mol of acid increases= 5+0.02 = 5.02

mol of conjguate left = 2 -0.02 = 1.98

substitute again

pH = 4.20 + log(1.98/5.02) = 3.79596

Q3

Ag2CO3 = 2Ag+ + CO3-2

Ksp = [Ag+]^2[CO3-2]

8.1*10^-12 = (2S)^2(S)

4*S^3 = 8.1*10^-12

S = ((8.1*10^-12)/(4))^(1/3)

S = 0.0001265 M

thisis 0.0001265 mol per liter

mol = MV = 0.0001265*0.25 = 0.000031625 mol

mass = mol*MW = 0.000031625*275.7453 = 0.00872 g = 8.72 mg of Ag2CO3