Question

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

a.$\left[{\text{H}}^{\text{+}}\right]{\text{= 1.0×10}}^{\text{- 7}}\text{M}$

b.$\left[{\text{H}}^{\text{+}}\right]{\text{= 8.3×10}}^{\text{- 16}}\text{M}$

c.$\left[{\text{H}}^{\text{+}}\right]\text{= 12}\text{M}$

d.$\left[{\text{H}}^{\text{+}}\right]\text{=}{\text{5.4×10}}^{\text{- 5}}\text{M}$

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 $\left[{\text{H}}^{\text{+}}\right]{\text{= 1.0×10}}^{\text{- 7}}\text{M}$(Neutral solution)

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

(b) For solution with $\left[{\text{H}}^{\text{+}}\right]{\text{= 8.3×10}}^{\text{- 16}}\text{M}$(Strongly Basic solution)

$\left[O{H}^{-}\right]=12.05,\text{pOH}=-1.08,\text{pH}=15.08$

(c) For solution with $\left[{\text{H}}^{\text{+}}\right]\text{= 12M}$(Strongly Acidic solution)

$\left[O{H}^{-}\right]=8.33\times {10}^{-16},\text{pOH}=15.05,\text{pH}=-1.05$

(d) For solution with $\left[{\text{H}}^{\text{+}}\right]{\text{= 5.4×10}}^{\text{- 5}}\text{M}$(Acidic solution)

$\left[O{H}^{-}\right]=1.85\times {10}^{-10},\text{pOH}=9.733,\text{pH}=4.267$

Step-by-step solution

Step-by-Step Solution Subpart (a) For solution of  H +   = 1.0×10 - 7M 

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

$\begin{array}{rcl}{\text{K}}_{\text{w}}& =& \left[H^{+}\right]\left[O{H}^{-}\right]\\ & =& 1\times {10}^{-14}\\ & & \end{array}$

Step 1: Concentration of $\left[{\text{OH}}^{\text{-}}\right]$

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

pH and pOH:

$\begin{array}{rcl}\text{pOH}& =& -\log \left[{\text{OH}}^{\text{-}}\right]\\ & =& -\log (1.0\times {10}^{-7})\\ & =& 7\end{array}$

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

As $\left[{\text{H}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]$, so the solution is neutral.

Subpart (b) For solution of$\left[{\text{H}}^{\text{+}}\right]\text{=}{\text{8.3×10}}^{\text{- 16}}\text{M}$

Step 1:Concentration of $\left[{\text{OH}}^{\text{-}}\right]$

$\left[O{H}^{-}\right]=\frac{K_w}{\left[H^{+}\right]}=\frac{1.0\times {10}^{-14}}{8.3\times {10}^{-16}}=12.05$

Step 2: pH and pOH:

$\begin{array}{rcl}\text{pOH}& =& -\log \left[{\text{OH}}^{\text{-}}\right]\\ & =& -\log (12.05)\\ & =& -1.08\\ & & \end{array}$

$\begin{array}{rcl}\text{pH}& =& 14-\text{pOH}\\ & =& 14-\text{( - 1.08)}\\ & =& 15.08\\ & & \end{array}$

As $\left[{\text{H}}^{\text{+}}\right]$ion concentration is very low so it is strongly basic solution.

Subpart (c) For solution of$\left[{\text{H}}^{\text{+}}\right]\text{=}\text{12}\text{M}$

Step 1:Concentration of $\left[{\text{OH}}^{\text{-}}\right]$

$\left[O{H}^{-}\right]=\frac{K_w}{\left[H^{+}\right]}=\frac{1.0\times {10}^{-14}}{12}=8.33\times {10}^{-16}$

Step 2: pH and pOH:

$\begin{array}{rcl}\text{pOH}& =& -\log \left[{\text{OH}}^{\text{-}}\right]\\ & =& -\log (8.33\times {10}^{-16})\\ & =& 15.05\\ & & \end{array}$

$\begin{array}{rcl}\text{pH}& =& 14-\text{pOH}\\ & =& 14-15.05\\ & =& -1.05\\ & & \end{array}$

As $\left[{\text{H}}^{\text{+}}\right]$ion concentration is very high, even more than 1M, so it is strongly acidic solution.

Subpart (c) For solution of$\left[{\text{H}}^{\text{+}}\right]\text{=}{\text{5.4×10}}^{-5}\text{M}$

Step 1:Concentration of $\left[{\text{OH}}^{\text{-}}\right]$

$\left[O{H}^{-}\right]=\frac{K_w}{\left[H^{+}\right]}=\frac{1.0\times {10}^{-14}}{5.4\times {10}^{-5}}=1.85\times {10}^{-10}$

Step 2: pH and pOH:

$\begin{array}{rcl}\text{pOH}& =& -\log \left[{\text{OH}}^{\text{-}}\right]\\ & =& -\log (1.85\times {10}^{-10})\\ & =& 9.733\\ & & \end{array}$

$\begin{array}{rcl}\text{pH}& =& 14-\text{pOH}\\ & =& 14-9.733\\ & =& 4.267\\ & & \end{array}$

As pH is less than 7 so it is an acidic solution.