Question

If $K_w$ for pure $H_{2}O$ at ${45}^{\circ }C$ is $4\times {10}^{-14}{M}^2$. The pH of water at this temperature is :

• A. 5.7
• B. 3.7
• C. 4.7
• D. 6.7

Answer 1

The correct option is D 6.7

We know that,
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
As water is neutral and for a neutral solution:
$\left[H^{+}\right]=\left[O{H}^{-}\right]\therefore K_w=[H^{+}{]}^2\because K_w=4\times {10}^{-14}\text{(Given)}\Rightarrow [H^{+}{]}^2=4\times {10}^{-14}\left[H^{+}\right]=\sqrt{4\times {10}^{-14}}=2\times {10}^{-7}pH=-lo{g}_{10}\left[H^{+}\right]\therefore pH=-lo{g}_{10}(2\times {10}^{-7})pH=-(0.3-7)pH=6.7$