1 g of a monobasic acid (HA) in 100 g water lowers the freezing point by 0.1 K. If 0.75 g of same acid requires 15 ml of 1 5 N NaOH solution for complete neutralization, then % degree of ionization of

Question

1 g of a monobasic acid (HA) in 100 g water lowers the freezing point by 0.1 K. If 0.75 g of same acid requires 15 ml of 1 5 N NaOH solution for complete neutralization, then % degree of ionization of acid in water is K f of H 2 O = 1 .86 K kg mol - 1

TestHub · Accepted Answer

Normal molecular weight of acid can be found as

Meq of acid = Meq of NaOH

$N_{1} V_{1} = N_{2} V_{2}; N_{1} V_{1} = M_{2} V_{2}$

$\frac{0 .75}{M_{acid}} \times 1000 = 15 \times \frac{1}{5}; M_{acid} = 250$

Observed molecular weight of acid can be formed as

$∆ T_{f} = K_{f} \frac{W_{acid}}{M_{acid}} \times \frac{1000}{W_{solvent}} \times i; 0 .1 = 1 .86 \times \frac{1}{250} \times \frac{1000}{100} \times (1 + α)$

$1 .34 = 1 + α$

$α = 0 .34$

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