Question

At $t^{\circ }C$, $K_w$ for water is $6.4\times 10^{-13}$. The $pH$ of water at $t^{\circ }C$ will be

• A. 6.91
• B. 5.90
• C. 6.09
• D. 5.15

Answer 1

Answer: (C)

$P{K}_w=-log{K}_w=-log(6.4\times 10^{-13}=12.194$
$PH=POH$ for neutral solution
$PH=\frac{12.194}{2}=6.097$