Chapter 16, Problem 3STP

Interpretation Introduction

Interpretation:The true statement about a solution in which $[{\text{OH}}^{\text{-}}]=4.8\times {10}^{-4}\text{M}$ 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 3STP

a. $[{\text{H}}^{\text{+}}]=2.1\times {10}^{-11}\text{M}$ and the solution is basic.

Explanation of Solution

Given information:

  $[{\text{OH}}^{\text{-}}]=4.8\times {10}^{-4}\text{M}$

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 $[{\text{OH}}^{\text{-}}]$

  $\begin{array}{c}[{\text{H}}^{+}]=\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{[OH}}^{\text{-}}\text{]}}\\ [{\text{H}}^{+}]=\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{4.8\times {10}^{-4}}\\ [{\text{H}}^{+}]=2.1\times {10}^{-11}\end{array}$

Since, $[{\text{H}}^{+}]=\text{2}{\text{.1×10}}^{\text{-11}}\text{M}$ , let us calculate the pH by following

  $\begin{array}{c}\text{pH}=-\text{log}[{\text{H}}^{+}]\\ =-\log [2.1\times {10}^{-11}]\\ =10.7\end{array}$

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

Conclusion

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