Question

___________has a $pH$ of 13.

• A. 0.1 M $MgC{l}_2$
• B. 0.1 M $HCl{O}_4$
• C. 0.1 M $N{H}_{4}OH$
• D. 0.1 M $KOH$
• E. 0.1 M $LiN{O}_3$

Answer 1

Answer: (D)

0.1 M $KOH$

We know that $pOH$$=$$-log\left[O{H}^{-}\right]$.

$KOH$$=$$K^{+}$$+$$O{H}^{-}$

Thus, $0.1M$ of $KOH$ will give $0.1M$ of $\left[O{H}^{-}\right]$

Let us calculate the $pOH$ of the system-

$pOH$$=$$-log\left[0.1\right]$, since concentration of $\left[O{H}^{-}\right]$ is $0.1M$

$pOH$$=$$-log$($\frac{1}{10}$)

$pOH$$=$$-log$(${10}^{-1}$)

That is, $pOH$$=$$+log\left(10\right)$, which is equal to $1$

Thus, $pOH$ of this solution is equal to $1$.

We also know that, $pOH$$+$$pH$$=$$14$

Hence, $pH$$=$$14$$-$$pOH$ and $pOH$$=$$1$

$pH$$=$$14$$-$$1$, which is equal to $13$.

Option D is the correct answer