Question

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

Short 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-by-step solution

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 ${\mathbf{OH}}^{-}$ .

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 $K_w=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$