Question

Equal volume of two solution having $pH=2$ and $pH=10$ are mixed together at ${90}^{o}C$. Then pH of resulting solution is:

[Take $K_w$ at ${90}^{o}C={10}^{-12}]$

• A. 2 + log 2
• B. 10 $-$ log 2
• C. 7
• D. 6

Answer 1

Answer: (D)

6

The ionic product of water at ${90}^{o}C$ is ${10}^{-12}$.

Hence $p{K}_w=12$. Thus the pH at the neutral point will be 6.

When equal volumes of solutions having pH 2 $\left(\right[H^{+}]=0.01M)$ and pH 10$\left(\right[O{H}^{-}]=0.01M)$ are added at ${90}^{o}C$, a neutral solution with pH 6 is obtained.