50 mL of normal sulphuric acid neutralises 10 mL of potash solution. Calculate the strength of potash in mol `L^(-) and g L^(-1)`.

Explanation

In volumetric titration of `H_2SO_4` with KOH solution
`V_1N_1 = V_2N_2 "(or)" 50 xx 1 = 10 xx N ("or ") N = 1.25 mol L^(-1)`
For KOH, n = 1. Hence strength of KOH = `1.25 mol L^(-1)`
Strength of given potash solution = `1.25 xx 56.1 = 70.125 g L^(-1)` .