Question

A) A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 MM NaOH. Calculate the pH after the addition of 17.0 mL of NaOH.

B) Calculate the molar solubility of lead thiocyanate in 1.00 M KSCN. Lead thiocyanate, Pb(SCN)2 has a Ksp value of 2.00×10−5.

C) Calculate the molar solubility of AgBr in 0.10 M NaBr solution.

D) Calculate the molar solubility of Ni(OH)2Ni(OH)2 when buffered at pH=pH= 10.2. The Ksp of nickel hydroxide =6.0×10−16 M

Answer 1

a)

pH of acidic buffer = pka + log(CH3COONa/CH3COOH)

pH of acidic buffer = ?

pka of CH3COOH =    4.74

no of mol of CH3COOH = 52*0.35 = 18.2 mmole

No of mol of CH3COONa = NaOH added = 17*0.4 = 6.8 mmole

pH = 4.74 + log(6.8/(18.2-6.8))

pH = 4.516

b) Pb(SCN)2(S) ----> Pb+2(aq) + 2 SCN-(aq)

Ksp of Pb(SCN)2 = s*(2s)^2

Ksp of Pb(SCN)2 = 2.0*10^-5

[SCN-] = 2S = 1.00 M

2.0*10^-5 = s*(1)^2

s = molar solubility of Pb(SCN)2 = 2*10^-5 M

c)

Ag+(aq) + Br-(aq) ----> AgBr(s)

Ksp of AgBr = [Ag+][Br-] = s*s

Ksp of AgBr = 1.7*10^-10

From NaBr, [Br-] = s = 0.1 M

1.7*10^-10 = s*0.1

s = molar solubility of AgBr = 1.7*10^-9 M

d) Ni(OH)2(s) <-----> Ni+2(aq) + 2OH-(aq)

Ksp of Ni(OH)2 = [Ni+2][OH-]^2

pH = 10.2 , [H+] = 10^-10.2 M = 6.31*10^-11 M

[OH-] = 1*10^-14/[H+] = (1*10^-14)/(6.31*10^-11)

[OH-] = 2s = 1.58*10^-4 M

Ksp of Ni(OH)2 = s*(2s)^2

6.0*10^-16 = s*(1.58*10^-4)

s = 3.8*10^-12 M

s = molar solubility of Ni(OH)2 = 3.8*10^-12 M.