Question

Calculate $\left[H^{+}\right]$andpHfor the following solutions:

(a)$\mathbf{0}\mathbf{.}\mathbf{010}\mathbf{M}\mathbf{}\mathbf{}{\mathbf{HNO}}_{\mathbf{3}}$

(b)$\mathbf{0}\mathbf{.}\mathbf{035}\mathbf{}\mathbf{M}\mathbf{}\mathbf{}\mathbf{KOH}$

(c)$\mathbf{0}\mathbf{.}\mathbf{030}\mathbf{M}\mathbf{}\mathbf{}\mathbf{HCI}$

(d)$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{M}\mathbf{}\mathbf{}\mathbf{HCI}$

(e) $\mathbf{0}\mathbf{.}\mathbf{010}\mathbf{}\mathbf{M}\left[{\left({\mathrm{CH}}_3\right)}_4{N}^{+}\right]{\mathbf{OH}}^{\mathbf{-}}$

Tetramethylammonium hydroxide

Short answer

(a)

$\left[{\mathrm{H}}^{+}\right]=0.010\mathrm{M},\mathrm{pH}=2.00$

(b)

$\left[{\mathrm{OH}}^{-}\right]=0.035\mathrm{M},\mathrm{pH}=12.54$

(c)

$\left[{\mathrm{H}}^{+}\right]=0.030\mathrm{M},\mathrm{pH}=1.52$

(d)

$\left[{\mathrm{H}}^{+}\right]=3.0\mathrm{M},\mathrm{pH}=-0.48$

(e)

$\left[{\mathrm{OH}}^{-}\right]=0.010\mathrm{M},\mathrm{pH}=12.00$

Step-by-step solution

Calculate the pH value of the given solutions

The pH of the solution can be calculated as:

$\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]$

Calculation of pH of 0.010 M  HNO3 

(a)

The concentration of the given solution is $0.010\mathrm{M}$.

${\mathrm{HNO}}_3\to {\mathrm{H}}^{+}+{\mathrm{NO}}_3^{‐}$

The hydrogen ion concentration is equal to the actual concentration of the solution.

Therefore, $\left[{\mathrm{H}}^{+}\right]=0.010\mathrm{M}$

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ =-\log \left(0.010\right) \\ =2.00 \end{gathered}$$

Calculation of pH of 0.035 M  KOH 

The concentration of the given solution is 0.035M .

$\mathrm{KOH}\to {\mathrm{K}}^{+}+{\mathrm{OH}}^{‐}$

The concentration of the hydroxyl ion is equal to the actual concentration of the solution.

Therefore, $\left[{\mathrm{OH}}^{-}\right]=0.035\mathrm{M}$

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\frac{\mathrm{K}}{\left[{\mathrm{OH}}^{-}\right]} \\ =\frac{1.0\times {10}^{-14}}{0.035\mathrm{M}} \\ =2.86\times {10}^{-13}\mathrm{M} \\ \mathrm{pH}=-\log \left(2.86\times {10}^{-13}\right) \\ =12.54 \end{gathered}$$

Calculation of pH of  0.030M  HCI

The concentration of the given solution is 0.030M

$\mathrm{HCI}\to {\mathrm{H}}^{+}+{\mathrm{CI}}^{-}$

The hydrogen ion concentration is equal to the actual concentration of the solution.

Therefore, $\left[{\mathrm{H}}^{+}\right]=0.030\mathrm{M}$

localid="1663390304057" $$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ =-\log \left(0.030\right) \\ =1.52 \end{gathered}$$

Calculation of pH of 3.0 M  HCI 

The concentration of the given solution is 3.0M .

$\mathrm{HCI}\to {\mathrm{H}}^{+}+{\mathrm{CI}}^{-}$

The hydrogen ion concentration is equal to the actual concentration of the solution.

Therefore, $\left[{\mathrm{H}}^{+}\right]=3.0\mathrm{M}$

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ =-\log \left(3.0\right) \\ =-0.48 \end{gathered}$$

Calculation of pH of 0.010M[CH34N+]OH- 

The concentration of the given solution is 0.010M

The concentration of hydroxyl ion is equal to the actual concentration of the solution.

Therefore, $\left[{\mathrm{OH}}^{-}\right]=0.010\mathrm{M}$

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{-}\right]} \\ =\frac{1.0\times {10}^{-14}}{0.010\mathrm{M}} \\ =1.0\times {10}^{-12}\mathrm{M} \\ \mathrm{pH}=-\log \left(1.0\times {10}^{-12}\right) \\ =12.00 \end{gathered}$$