Question

Calculate pH of 0.00424 M aqueous solution of KOH.Solve the given problem.

Answer 1

Final answer: 11.63

Steps:

1) We have,
Since,KOH is strong base .

$[OH^-] = 0.00424 = 424 * 10^{-5}$

2) We know,

pH = 14- pOH

$pH = 14-(-\log([OH^-])) \\ = 14+\log([OH^-] ) \\ = 14+ \log(424*10^{-5}) \\ = 14-5+\log(424) \\ = 9+ 2.63 \\ =11.63$

Hence, pH of 0.00424M solution of KOH is 11.63