Question

Calculate the $\left[{\text{H}}^{\text{+}}\right]$of each of the following solutions at$\mathbf{25}°\mathbf{C}$ . Identify each solution as neutral, acidic, or basic.

a.$\left[{\text{OH}}^{-}\right]=1.5\text{M}$

b.$\left[{\text{OH}}^{-}\right]=3.6\times {10}^{-5}\text{M}$

c.$\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-7}\text{M}$

d.$\left[{\text{OH}}^{-}\right]=7.3\times {10}^{-4}\text{M}$

Also, calculate the pH and pOH of each of these solutions.

Short answer

$\text{pH}\text{+ pOH =}\text{14}$at $\mathbf{25}°\mathbf{C}$.

a. For solution with $\left[{\text{OH}}^{\text{-}}\right]\text{= 1.5 M}$(Basic Solution)

${\text{[H}}^{\text{+}}\text{]}\text{=}{\text{6.67×10}}^{\text{- 13}}\text{,}\text{pH}\text{= 12.176,}\text{pOH}\text{= 1.824}$

b. For solution with $\left[{\text{OH}}^{\text{-}}\right]\text{= 3.6}\times {\text{10}}^{-15}\text{M}$(Strongly acidic solution)

${\text{[H}}^{\text{+}}\text{]}\text{=}\text{2.77,pH}\text{=}\text{- 0.442,}\text{pOH}\text{= 14.442}$

c. For solution with $\left[{\text{OH}}^{\text{-}}\right]\text{= 1.0}\times {\text{10}}^{-7}\text{M}$(Neutral solution)

${\text{[H}}^{\text{+}}\text{]}\text{=}{\text{1.0×10}}^{\text{- 7}}\text{,pH}\text{=}\text{7,}\text{pOH}\text{=}\text{7}$

For solution with $\left[{\text{OH}}^{\text{-}}\right]\text{= 7.3}\times {\text{10}}^{-4}\text{M}$(Basic solution)

${\text{[H}}^{\text{+}}\text{]}{\text{= 1.37×10}}^{\text{- 11}}\text{,}\text{pH}\text{=}\text{10.86,}\text{pOH}\text{=}\text{3.14}$

Step-by-step solution

Subpart (a) Step 1: Calculation of pH for solution of  OH-=1.5 M. 

For a solution at $\mathbf{25}°\mathbf{C}$,

$\begin{array}{rcl}& & {\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ & & {\text{= 1×10}}^{\text{- 14}}\\ & & \end{array}$

Therefore, the concentration of $\left[{\text{H}}^{\text{+}}\right]$

${\text{[H}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{[OH}}^{\text{-}}\text{]}}\text{=}\frac{{\displaystyle {\text{1.0×10}}^{\text{- 14}}}}{{\displaystyle \text{1.5}}}{\text{= 6.67×10}}^{\text{- 13}}$

The pH, therefore,

$\begin{array}{rcl}\text{pH}& =& -\log {\text{[H}}^{\text{+}}\text{]}\\ & =& -\log (6.67\times {10}^{-13})\\ & =& 12.176\\ & & \end{array}$

As the pH of the solution is more than 7 so its basic solution.

Calculation of pOH for solution of  OH-=1.5 M

$\begin{array}{rcl}\text{pOH}& =& 14-\text{pH}\\ & =& 14-12.176\\ & =& 1.824\\ & & \end{array}$

Subpart (b)

Step 1: Calculation of pH for solution of $\left[{\mathbf{OH}}^{-}\right]=3.6\times {10}^{-15}\mathbf{M}$

Concentration of$\left[{\text{H}}^{\text{+}}\right]$

$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{1.0\times {10}^{-14}}{3.6\times {10}^{-15}}=2.77$

The pH, therefore,

$\begin{array}{rcl}\text{pH}& =& -\log {\text{[H}}^{\text{+}}\text{]}\\ & =& -\log (2.77)\\ & =& -0.442\\ & & \end{array}$

As [H⁺] ion concentration is very high (even more than 1 M) so it is strongly acidic solution.

Step 2: Calculation of pOH for solutionof$\left[{\mathbf{OH}}^{-}\right]=3.6\times {10}^{-15}\mathbf{M}$

$\begin{array}{rcl}\text{pOH}& =& 14-\text{pH}\\ & =& 14-(-0.442)\\ & =& 14.442\\ & & \end{array}$

Subpart (c)

Step 1: Calculation of pH for solution of $\left[{\mathbf{OH}}^{-}\right]=1.0\times {10}^{-7}\mathbf{M}$

Concentration of$\left[{\text{H}}^{\text{+}}\right]$

$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-7}}=1.0\times {10}^{-7}$

The pH, therefore,

$\begin{array}{rcl}\text{pH}& =& -\log {\text{[H}}^{\text{+}}\text{]}\\ & =& -\log (1.0\times {10}^{-7})\\ & =& 7\\ & & \end{array}$

As [H⁺] = [OH^-], so the solution is neutral.

Step 2:Calculation of pOH for solution of$\left[{\mathbf{OH}}^{-}\right]=1.0\times {10}^{-7}\mathbf{M}$

$\begin{array}{rcl}\text{pOH}& =& 14-\text{pH}\\ & =& 14-7\\ & =& 7\\ & & \end{array}$

Subpart (d)

Step 1: Calculation of pH for solution of $\left[{\mathbf{OH}}^{-}\right]=7.3\times {10}^{-4}\mathbf{M}$

Concentration of $\left[{\text{H}}^{\text{+}}\right]$

$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{1.0\times {10}^{-14}}{7.3\times {10}^{-4}}=1.37\times {10}^{-11}$

The pH, therefore,

$\begin{array}{rcl}\text{pH}& =& -\log {\text{[H}}^{\text{+}}\text{]}\\ & =& -\log (1.37\times {10}^{-11})\\ & =& 10.86\\ & & \end{array}$

As pH is more than 7 so it is basic solution.

Step 2:Calculation of pOH for solution of$\left[{\mathbf{OH}}^{-}\right]=7.3\times {10}^{-4}\mathbf{M}$

$\begin{array}{rcl}\text{pOH}& =& 14-\text{pH}\\ & =& 14-10.86\\ & =& 3.14\\ & & \end{array}$