Question

Ph of solution at 25c which has twice as many hydroxide ion in pure water at 25c will be

Answer 1

Answer:The pH of the solution is 7.31.

Solution:

Hydroxide ion concentration in pure water at 25$^oC$ :

$K_w=1\times 10^{-14}=[OH^-][H^+]$

Since, pH of water is neutral which means that $[OH^-]=[H^+]$

$1\times 10^{-14}=[OH^-][OH^-]$

$[OH^-_{water}]=1\times 10^{-7}mol/L$

The pH of solution at 25$^oC$ which has twice as many hydroxide ion in pure water at 25$^oC$  will be:

$[OH^-_{solution}]=2\times [OH^-_{water}]=2\times 10^{-7}mol/L$

$pOH=-log[OH^-_{solution}]=-log(2\times 10^{-7})=6.69$

$pH=14-pOH=14-6.69=7.31$

The pH of the solution is 7.31.

Answer 2

Answer:

7.301

Explanation:

see attachment