Question

The ph of a sample of koh solution is 12.3979 the weight of solid koh of 70% pure required to prepare 2.5 lit of this solution is

Answer 1

5 g of KOH is of 70% pure is required to prepare 2.5 lit of this solution.

Explanation:

POH = 14 − 12.3979 = 1.6021

N of [OH−] = 2.5 × 10^−2 N

∴N = w / eq . wt' × 1000 / V(ml)

2.5 × 10^−2 = w / 56 × 12.5(l)

W = 56 × 2.5 × 2.5 × 10^−2 = 3.5 g

70% pure KOH means.

100 g  impure sub →70 g KOH

? → 3.5 g KOH

5g

Thus 5 g of KOH is of 70% pure required to prepare 2.5 lit of this solution.

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