Chapter 16, Problem 4STP

Interpretation Introduction

Interpretation: The pH of a solution in which ${\text{[OH}}^{\text{-}}\text{]}=\text{2}{\text{.3×10}}^{\text{-3}}\text{M}$ needs to be determined.

Concept introduction:Ion-product constant: An ion product constant is basically the same thing as an equilibrium constant.

  ${\text{K}}_{\text{w}}=[{\text{H}}^{\text{+}}][{\text{OH}}^{\text{-}}]=1.0\times {10}^{-14}$

Answer to Problem 4STP

D. 11.36

Explanation of Solution

Given information:

  ${\text{[OH}}^{\text{-}}\text{]}=\text{2}{\text{.3×10}}^{\text{-3}}\text{M}$

Let us calculate $[{\text{H}}^{+}]$ from the ${\text{K}}_{\text{w}}$ expression,

  ${\text{K}}_{\text{w}}=[{\text{H}}^{\text{+}}][{\text{OH}}^{\text{-}}]=1.0\times {10}^{-14}$

for $[{\text{H}}^{+}]$ by dividing both sides by $[{\text{OH}}^{\text{-}}]$

  $\begin{array}{c}[{\text{H}}^{+}]=\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{[OH}}^{\text{-}}\text{]}}\\ [{\text{H}}^{+}]=\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{2.3\times {10}^{-3}}\\ [{\text{H}}^{+}]=4.3\times {10}^{-12}\end{array}$

Since, $[{\text{H}}^{+}]=4.{\text{3×10}}^{\text{-12}}\text{M}$ , let us calculate the pH by following

  $\begin{array}{c}\text{pH}=-\text{log}[{\text{H}}^{+}]\\ =-\log [4.3\times {10}^{-12}]\\ =11.36\end{array}$

Since, $\text{pH}<7$ , the solution is basic.

Conclusion

The pH scale provides a compact way to represent solution acidity.