Ka and Kb Video Lessons

Concept:

# 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:

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

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###### 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