Question

At $25,\left[H_3{O}^{+}\right]=1\times {10}^{-7}M$ in water, hence, $K_a$ is:

• A. $1\times {10}^{-14}$
• B. $5.55\times {10}^{-15}$
• C. $1.8\times {10}^{-16}$
• D. $55.5\times {10}^{-10}$

Answer 1

The correct option is A $1\times {10}^{-14}$

Given that $\left[H_3{O}^{+}\right]=1\times {10}^{-7}M$

$k_a=?$

$HA+H_{2}O\rightleftharpoons H_3{O}^{+}+A^{-}$

$k_a=\frac{\left[H_3{O}^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

$\left[H_3{O}^{+}\right]=\left[A^{-}\right]$

$k_a=\frac{1\times {10}^{-7}\times 1\times {10}^{-7}}{1}$

$k_a={10}^{-14}$

Hence, the correct option is $\text{A}$