Question

Calculate the pH of a [0.000765) M solution of KOH

Answer 1

Answer:

pH=10.88

Explanation:

Hello,

In this case, since potassium hydroxide is completely dissociated as shown below:

$KOH\rightarrow K^++OH^-$

For which we understand it is a base, more specifically, a strong base; it means that the concentration of the OH⁻ equals the concentration of the potassium hydroxide, that is 0.000765M, for that reason we can directly compute the pOH:

$pOH=-log([OH^-])=-log(0.000765)=3.12$

Finally, since the pOH and the pH are related by:

$pOH+pH=14$

The pH turns out:

$pH=14-3.12\\pH=10.88$

Best regards.