Question

1) The Ka of a monoprotic weak acid is 2.47

Answer 1

HA <--> H^+1 + A^-1
Ka = [H+][A-]/[HA]
2.47x10^-3 = [H+][A-]/(0.169 - [H+])
[H+] = [A-]= X
2.47x10^-3 = X^2/(0.164 - X)
X will not be << 0.164 so you can't assume that 0.164 - X = 0.164
4.0508x10^-4 - 2.47x10^-3 X = X^2
X^2 + 2.47x10^-3 X - 4.0508x10^-4 = 0
Solving for X yields X = 0.01893
0.01893 / 0.164 = 0.115
ANSWER: 11.5% ionized

Use the Henderson-Hasselbalch equation: pOH = pKb + log[BH^+]/[B], where BH^+ represents the protonated amine (conjugate base) and B represents the amine (weak base).

For a monoprotic acid or base the expression: pKa + pKb = pKw, and pH + pOH = pKw are helpful.

pOH = 14.00 - 9.386 = 4.614
pKb = -log(6.524 x 10^-5) = 4.185

4.614= 4.185 + log[BH^+]/[B] = 0.429

You need to find [BH^+]/[B] so take the antilog of both sides.

Antilog(log [BH^+]/[B]) = [BH^+]/[B]
antilog 0.429 = 10^0.429 = 2.685

[BH^+]/[B] = 2.685

[BH^+] = 2.685[B]

% BH^+ protonated = {[BH^+]/([B] + [BH^+])}*100% = {2.685[B]/([B] + 2.685[B]}*100%
% BH^+ protonated = (2.685/3.685)*100% = 72.87% protonated

3

NH4Cl is a salt of weak base NH3 and strong acid HCl ...and pH of such salt solution is given by :-

pH = 1/2 [ pKw -pKb - log c ]

where pKw = -log 10^-14 = 14
pKb = -log Kb = -log (1.8 X 10^-5) = 4.745

log c = log 0.036 = -1.44

pH = 1/2 [ 14 - 4.745 - (-1.44) ] = 1/2 [ 14 - 4.745 + 1.44 ] = 10.9655/2 = 5.4827