Question

5. At 20°C, the water autoionization constant, Kw, is 6.8 ´ 10–15. What is the H3O+ concentration in neutral water at this temperature? A. 6.8 × 10–7 M B. 3.4 × 10–15 M C. 6.8 × 10–15 M D. 8.2 × 10–8 M E. 1.0 × 10–7 M

Answer 1

Explanation:

Let us assume that the concentration of [$OH^{-}$ and $H^{+}$ is equal to x. Then expression for $K_{w}$ for the given reaction is as follows.

          $K_{w} = [OH^{-}][H^{+}]$

          $K_{w} = x^{2}$

      $6.8 \times 10^{-15} = x^{2}$

Now, we will take square root on both the sides as follows.

          $\sqrt{6.8 \times 10^{-15}} = \sqrt{x^{2}}$

          $[H^{+}] = 8.2 \times 10^{-8}$ M

Thus, we can conclude that the $H_{3}O^{+}$ concentration in neutral water at this temperature is $8.2 \times 10^{-8}$ M.

Answer 2

Answer: The concentration of $H_3O^+$ in neutral water is $8.2\times 10^{-8}M$

Explanation:

The chemical equation for the ionization of water follows:

$2H_2O\rightleftharpoons H_3O^++OH^-$

The expression of $K_w$ for above equation, we get:

$K_w=[H_3O^+]\times [OH^-]$

We are given:

$K_w=6.8\times 10^{-15}$

$[H^+]=[OH^-]=x$

Putting values in above equation, we get:  

$6.8\times 10^{-15}=x\times x\\\\x=8.2\times 10^{-8}M$

Hence, the concentration of $H_3O^+$ in neutral water is $8.2\times 10^{-8}M$