Question

2055 Q. No. 10^-2

Answer 1

Answer:

11

Explanation:

Moles of KOH = $10^{-2}$

Volume of water = 10 liters

Concentration of KOH is given by

$[KOH]=\dfrac{10^{-2}}{10}\\\Rightarrow [KOH]=10^{-3}\ \text{M}$

$[KOH]$ is strong base so we have the following relation

$[KOH]=[OH^{-}]=10^{-3}\ \text{M}$

$pOH=-\log [OH^{-}]=-\log10^{-3}$

$\Rightarrow pH=14-3=11$

So, pH of the solution is 11