An aqueous solution of an acid is so weak that it can be assumed to be pratically unionised, boiled at ${100.4}^{o} C$ . 25 mL of this solution was neutralised by 38.5 mL of 1 N solution of NaOH. Calculate basicity of the acid if $K_{b}$ for water is $0.52 K m o l^{- 1}$ kg. Assume molality equal to molarity.

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Solution

Meq. of acid in 25 mL $= 38.5 \times 1$
$∴$ Meq. of acid in 1000 mL $= \frac{38.5 \times 1 \times 1000}{25} = 1540$
$∴$ Normality of acid $= \frac{1540}{1000} = 1.54$
$∴$ Molarity of acid $= \frac{1.54}{n}$ (where, n is basicity of acid)
Also, $m o l a l i t y = \frac{Δ T}{K_{f}} = \frac{0.4}{0.52} = m o l a r i t y$ (given)
$∴ \frac{0.4}{0.52} = \frac{1.54}{n}$
$∴ n = 2$
Thus, the acid is dibasic.

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An aqueous solution of an acid is so weak that it can be assumed to be pratically unionised, boiled at 100.4oC. 25 mL of this solution was neutralised by 38.5 mL of 1 N solution of NaOH. Calculate basicity of the acid if Kb for water is 0.52Kmol−1 kg. Assume molality equal to molarity.

Meq. of acid in 25 mL =38.5×1∴ Meq. of acid in 1000 mL =38.5×1×100025=1540∴ Normality of acid =15401000=1.54∴ Molarity of acid =1.54n (where, n is basicity of acid)Also, molality=ΔTKf=0.40.52=molarity (given)∴0.40.52=1.54n∴n=2Thus, the acid is dibasic.