Question

The ionic product of water is found to be $1\times {10}^{-12}{mole}^2/l^2$ at ${60}^{o}C$. What would be the nature of solution with pH$=7$?

• A. Basic
• B. Acidic
• C. Neutral
• D. Cannot be determined

Answer 1

Answer: (A)

Basic

In pure water at ${60}^{o}C$ temperature the $Kw$ value:

$\left[H^{+}\right]\left[O{H}^{-}\right]=1.00\times {10}^{-12}$

But in pure water, the $\left[H^{+}\right]=\left[O{H}^{-}\right].$ can replace the $\left[O{H}^{-}\right]$ term in the $Kw$ expression by another $\left[H^{+}\right]$.

${\left[H^{+}\right]}^2=1.00\times {10}^{-12}$

Taking the square root of each side gives:

$\left[H^{+}\right]=1.00\times {10}^{-6}$

Converting that into pH:

$pH=-lo{g}_{10}\left[H^{+}\right]$

$pH=6$

At $pH=6$ solution will be neutral.At $pH>6$ solution is basic and $pH<6$ solution is acidic.