Question

In a titration of 40.55 mL of 0.3137 M nitrous acid with 0.3137 M aqueous sodium hydroxide, what is the pH of the solution when 40.55 mL of the base have been added?

Answer 1

i) Determine the number of moles of total no of moles of Nitrous acid present

we know that molarity is defined as the no of moles of solute per liter of solution

Therefore

No of moles of Nitrous acid = (0.3137mol/1000ml)×40.55ml = 0.01272mol

ii) Determine the number of moles of NaOH needed to neutralize 0.01272 moles of Nitrous acid

The reaction between HNO2 and NaOH is 1:1molar reaction

HNO2(aq) + NaOH(aq) -------> NaNO2 + H2O

Therefore,

0.01272 moles of HNO2 require 0.01272 moles of NaOH

iii) Determine the number of moles of NaOH added

as per the similar calculation above

Number of moles of NaOH added = (0.3137mol/1000ml)× 40.55ml = 0.01272

Therefore,

Therefore,

40.55ml equivalance point

iv) Determine the concentration of NO2^- at equivalence point

At equivalence point all the HNO2 converted to NO2^-

Total volume at equivalence point = 40.55ml + 40.55ml = 81.1ml

[NO2^-] = (0.01272mol/81.1ml) ×1000ml = 0.1568M

v) Determine the equillibrium concentration of OH^-

NO2^- is partly hydrolysed by H2O

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

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

K_b = K_w/K_a

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

K_a = dissociation constant of HNO2 , 4.00×10^-4

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

at equillibrium

[NO2^-] = 0.1568 - x

[HNO2] = x

[OH^-] = x

Therefore,

x²/(0.1568 - x) = 2.5×10^-11

solving for x

x = 1.9799×10^-6

Therefore,

[OH^-] = 1.9799×10^-6M

vi) Determine pOH

pOH = -log [OH^-]

pOH = -log(1.9799×10^-6)

pOH = 5.70

vii) Determine pH from the relation between pH and pOH

pH + pOH = 14

pH = 14 -pOH

pH = 14 - 5.70

pH = 8.30