Question

In a titration of 35.61 mL of 0.3889 M nitrous acid with 0.3889 M aqueous sodium hydroxide, what is the pH of the solution when 35.61 mL of the base have been added?

Answer 1

pH = 8.34

Explanation

i) The reaction between HNO2 and NaOH is as follows

HNO2 + OH^- ---------> H2O + NO2^-

This 1:1 reaction

V1M1 = V2 M2

V1 = 35.61ml

M1 = 0.3889M

M2 = 0.3889M

V2 = M1×V1/M2

V2 = 35.61ml × 0.3889M / 0.3889M = 35.61ml

Therefore,

Equivalence point = 35.61ml

ii) At equivalence point all the HNO2 converted into NO2^-

Total volume at euivalence point = 35.61ml + 35.61ml = 71.22ml

Dilution of [NO2^-] = 71.22ml /35.61ml = 2

[NO2^-] at equivalence point = 0.3889M/2 = 0.19445M

iii) NO2^- is partly hydrolysed by water

NO2^- (aq) + H2O(l) ------> HNO2(aq) + OH^-(aq)

K_b= [HNO2][OH^-] /[ NO2^- ]

K_b = K_w/K_a

where, K_w = Ionic product of water, 1.00×10^-14

K_a = dissociation constant of nitrous acid, 4.00×10^-4

K_b = 1.00×10^-14/4.00×10^-4 = 2.5×10^-11

ICE table for this equillibrium

concentration	[NO2-]	[HNO2]	[OH-]
Initial concentration	0.19445	0	0
Change in concentration	-x	+x	+x
Equillibrium concentration	0.19445 -x	x	x

Therefore,

x²/(0.19445 -x) = 2.5×10^-11

we can assume 0.19445 - x ~ 0.19445 because x is small value

x²/0.19445 = 2.5×10^-11

x² = 4.86×10^-12

x = 2.20×10^-6

[OH^-] = 2.20×10^-6M

pOH = -log[OH^-]

pOH = 5.66

iv) Relation between pH and pOH is as follows

pH + pOH = 14

pH = 14 - 5.66

pH = 8.34