Question

To neutralise completely 20 ml of 0.1 m aqueous solution of phosphorous acid (h3po3), the value of 0.1 m aqueous koh solution required is

Answer 1

Hey !!

H₃PO₃ is dibasic while KOH is monoacidic

           na Ma Va = nb Mb Vb

 = 2 × 0.1 × 20 = 1 × 0.1 ×  Vb (OR) Vb = 40 mL

Hope it helps you !!

Answer 2

The volume of the $KOH$ added will be 40 ml.

Explanation:

According to neutralization law:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of $H_3PO_3$ = 2

$n_2$ = acidity of $KOH$ = 1

$M_1$ = concentration of $H_3PO_3$ = 0.1 M

$M_2$ = concentration of KOH = 0.1 M

$V_1$ = volume of $H_3PO_3$ = 20 ml

$V_2$ = volume of KOH = ?

Now put all the given values in the above law, we get the volume of $KOH$ added.

$(2\times 0.1M\times 20)=(1\times 0.1M\times V_2)$

By solving the terms, we get :

$V_2=40ml$

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