Question

The value of the ion product constant of water, $\left(K_w\right)$ at ${60}^{\circ }C$ is $9.6\times {10}^{-14}{M}^2$. What is the $\left[H_3{O}^{+}\right]$ of a neutral aqueous solution at ${60}^{\circ }C$ and an aqueous solution with a $pH=7.0$ at ${60}^{\circ }C$ are respectively?

• A. $3.1\times {10}^{-8}$, acidic
• B. $3.1\times {10}^{-7}$, neutral
• C. $3.1\times {10}^{-8}$, basic
• D. $3.1\times {10}^{-7}$, basic

Answer 1

The correct option is D $3.1\times {10}^{-7}$, basic

Solution:- (D) $3.1\times {10}^{-7}$, basic

Given:-

$K_w=9.6\times {10}^{-14}{M}^2$

As we know that,

for a neutral solution,

$\left[{H_{3}O}^{+}\right]=\left[{OH}^{-}\right]$

Now,

$K_w=\left[{H_{3}O}^{+}\right]\times \left[{OH}^{-}\right]$

$\Rightarrow K_w={\left[{H_{3}O}^{+}\right]}^2(\because \left[{H_{3}O}^{+}\right]=\left[{OH}^{-}\right])$

$\Rightarrow {\left[{H_{3}O}^{+}\right]}^2=9.6\times {10}^{-14}$

$\Rightarrow \left[{H_{3}O}^{+}\right]=\sqrt{9.6\times {10}^{-14}}\approx 3.1\times {10}^{-7}$

Hence at $60℃$, for a neutral aqueous solution, the $\left[{H_{3}O}^{+}\right]$ is $3.1\times {10}^{-7}$.

Now,

pH of the neutral solution-

$pH=-\log \left[{H_{3}O}^{+}\right]$

$pH=-\log (3.1\times {10}^7)$

$\Rightarrow pH=(7-\log \left(3.1\right))=6.51$

Since pH of a neutral solution is less than $7$, the aqueous solution with a $pH=7$ is basic in nature.