Question

1. What is the pH change of a 0.210 M solution of citric acid (pKa=4.77) if citrate is added to a concentration of 0.125 M with no change in volume?

2. What is the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2?

Answer 1

Ans 1

Dissociation of citric acid with ICE TABLE

C?H?O? = H⁺ + C?H?O?^-

I 0.210

C - x +x +x

E (0.210-x) x x

Equilibrium constant expression of the reaction

Ka = [C?H?O?-] [H+] /[C?H?O?]

pKa = 4.77

Ka = 10^-4.77 = 1.70*10^-5

Ka = x²/(0.210 - x)

Ka = 1.70*10^-5

x << 0.210

0.210 - x ? 0.210

Ka = x²/0.210

x = [H+] = ?[1.7*10^-5*0.210] = 0.00188

pH = - log [H+] = - log (0.00188)

pH = 2.73

Now add 0.125 M citrate

[C?H?O?^-] = 0.125

[ C?H?O? ] = 0.210

From the Henderson Hasselbalch equation

pH = pKa + log[C?H?O?^-]/[C?H?O? ]

= 4.77 + log(0.125/0.210)

= 4.77 - 0.23

= 4.54

Change in pH = 4.54 - 2.73 = 1.81