Question

$K_{w}for{H}_{2}Ois9.62\times {10}^{-14}at{60}^{\circ }C$
The pH of water at this temperature is :

• A. 6.5
• B. 7.5
• C. 7
• D. 6

Answer 1

The correct option is A 6.5

We know that,
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
As water is neutral and for a neutral solution:
$\left[H^{+}\right]=\left[O{H}^{-}\right]\therefore K_w=[H^{+}{]}^2\because K_w=9.62\times {10}^{-14}(Given)\Rightarrow [H^{+}{]}^2=9.62\times {10}^{-14}\left[H^{+}\right]=\sqrt{9.62\times {10}^{-14}}=31\times {10}^{-8}pH=-lo{g}_{10}\left[H^{+}\right]\therefore pH=-lo{g}_{10}(31\times {10}^{-8})pH=6.5$