Q18.20P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the difference between  ${\mathbf{\text{K}}}_{\mathbf{\text{c}}}$ and ${\mathbf{\text{K}}}_{\mathbf{\text{w}}}$ for the autoionization of water?

Step-by-Step Solution

Answer

The difference is:

$K_c=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]}{{\left[{\text{H}}_2\text{O}\right]}^2}\text{ and }{K}_w=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$ 

Step 1: Definition of concept

The autoionization of water refers to the reaction of water molecules to form two ions, the hydronium ion, which is $\mathbf{H}\mathbf{3}{\mathbf{O}}^{+}$  , and the hydroxide ion, which is ${\mathbf{OH}}^{-}$ .

Step 2: Explanation

The equilibrium constant of water is $Kc$

 $$\begin{gathered} {\text{H}}_2\text{O }+{\text{ H}}_2\text{O }\rightleftharpoons {\text{ H}}_3{\text{O}}^{+}+{\text{ OH}}^{-} \\ {K}_c=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[O{\text{H}}^{-}\right]}{{\left[{\text{H}}_2\text{O}\right]}^2} \end{gathered}$$

We move it to the other side since the concentration  ${\text{H}}_2\text{O}$ is constant.

$K_c{\left[H_{2}O\right]}^2=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]$ 

After that, we create a new variable called $K_c{\left[H_{2}O\right]}^2$ , which is the $K_w$ . So there you have it.

 $$\begin{gathered} K_w=K_c{\left[H_{2}O\right]}^2 \\ =\left[H_3{O}^{+}\right]\left[O{H}^{-}\right] \end{gathered}$$

 Therefore, The difference between $K_c$ and $K_w$ is that  $K_c\left[{\text{H}}_3{\text{O}}^{+}\right]$

  $K_c=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]}{{\left[{\text{H}}_2\text{O}\right]}^2}$ and  $K_w=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$