# Problem: At 40 °C, the value of Kw = 2.92 × 10–14. An acidic solution at 40 °C will therefore have:A. [H3O+] &gt; [OH–] and [H3O+] = 1.70 × 10–7 MB. [H3O+] &gt; [OH–] and [H3O+] &gt; 1.70 × 10–7 MC [H3O+] &gt; [OH–] and [H3O+] = 2.96 × 10–7 MD. [H3O+] = [OH–] and [OH–] = 1.70 × 10–7 ME. [H3O+] &lt; [OH–] and [OH–] &gt; 1.70 × 10–7 M

###### FREE Expert Solution

We’re being asked to determine [H3O+] and [OH-] of an acidic solution a40 °C if the value of Kw is 2.92 × 10–14.

Recall that Kw is the autoionization constant of water, and is given by:

$\overline{){{\mathbf{K}}}_{{\mathbf{w}}}{\mathbf{=}}\left[{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\right]\left[{\mathbf{OH}}^{\mathbf{-}}\right]}$

###### Problem Details

At 40 °C, the value of Kw = 2.92 × 10–14. An acidic solution at 40 °C will therefore have:

A. [H3O+] > [OH] and [H3O+] = 1.70 × 10–7 M

B. [H3O+] > [OH] and [H3O+] > 1.70 × 10–7 M

C [H3O+] > [OH] and [H3O+] = 2.96 × 10–7 M

D. [H3O+] = [OH] and [OH] = 1.70 × 10–7 M

E. [H3O+] < [OH] and [OH] > 1.70 × 10–7 M

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Ka and Kb concept. You can view video lessons to learn Ka and Kb. Or if you need more Ka and Kb practice, you can also practice Ka and Kb practice problems.