Question

Calculate the pH of the following solutions.

a. 0.10 M NaOH

b. 1.0 × 10^-10 M NaOH

c. 2.0 M NaOH

Short answer

pH for different solutions is given as

(a) For 0.10 M NaOH,

(b) For 1.0 × 10^-10 M NaOH,

(c) For 2 M NaOH,

Step-by-step solution

pH of 0.10 M NaOH

NaOH is a very strong base so it dissociates almost completely. So,

[NaOH] = [OH^-]

$\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left[0.1\right]=1$

At 25°C,

$$\begin{gathered} \mathrm{pH}+\mathrm{pOH}=14 \\ \mathrm{pOH}=14-\mathrm{pH}=13 \end{gathered}$$

pH of 1.0 × 10-10 M NaOH

As for, [NaOH] = [OH^-] = 1.0 × 10^-10 M

This value is low as compared to [OH^-] comes from water, so [OH^-] value of water cannot be ignored.

$$\begin{gathered} \mathrm{Total}\left[{\mathrm{OH}}^{-}\right]=\left[{\mathrm{OH}}^{-}\right]\mathrm{from}\mathrm{NaOH}+\left[{\mathrm{OH}}^{-}\right]\mathrm{from}{\mathrm{H}}_2\mathrm{O} \\ =1.0\times {10}^{-10}+1.0\times {10}^{-7}=\left(1+0.001\right)\times {10}^{-7} \end{gathered}$$

$$\begin{gathered} \mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left[(1.001)\times {10}^{-7}\right]=6.99 \\ \mathrm{pH}=14-6.99=7.01 \end{gathered}$$

pH of 2 M NaOH

For, [NaOH] = [OH^-]

$\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left[2\right]=-0.3010$

At 25°C,

$\mathrm{pH}=14-\mathrm{pOH}=14-(-0.3010)=14.3010$