Q.10.92 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]and\left[{\mathrm{OH}}^{-}\right]$ for a solution with the following pH values:

a. 10.0

b. 5.0

c. 7.00

d. 6.5

Step-by-Step Solution

Answer

Part a.$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0\times {10}^{-10}\mathrm{M},\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-4}\mathrm{M}$

Part b.$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0\times {10}^{-5}\mathrm{M},\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-9}\mathrm{M}$

Part c.$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0\times {10}^{-}7\mathrm{M},\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-7}\mathrm{M}$

Part d.$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.51\times {10}^{-2}\mathrm{M},\left[{\mathrm{OH}}^{-}\right]=6.6\times {10}^{-13}\mathrm{M}$

Step 1: Given Information (Part a)

As a solution gets more acidic as H+ increases, the pH diminishes. As a solution gets more basic (higher [OH-]), the pH increases. As the pH of a solution diminishes by one pH unit, the concentration of H+ increases by ten times

Step 2: Explanation Part (a)

Given, pH = $10$

We know, pH = -log

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}}$ and $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14}$

$$\begin{gathered} \Rightarrow \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-10} \\ =1.0\times {10}^{-10}\mathrm{M} \end{gathered}$$

$\left[{\mathrm{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{10}}=1\times {10}^{-4}\mathrm{M}$

Step 3: Explanation Part (b)

Given pH = $5$

We know, pH = -log

$\begin{array}{r}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}}\text{ and }\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14}\end{array}$

$\Rightarrow \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-5}$

$=1.0\times {10}^{-5}\mathrm{M}$

$\begin{array}{r}\left[{\mathrm{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-5}}=1\times {10}^{-9}\mathrm{M}\end{array}$

Step 4: Explanation Part (c)

Given pH = $7$

We know, pH = -log
$\begin{array}{r}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}}\text{ and }\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14}\end{array}$

$\Rightarrow \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-7}$$=1.0\times {10}^{-7}\mathrm{M}$

$\begin{array}{r}\left[{\mathrm{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-7}}=1\times {10}^{-7}\mathrm{M}\end{array}$

Step 5: Explanation Part (d)

Given pH = $1.82$

We know, pH = -log

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}}\text{ and }\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14}$

$\Rightarrow \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-1.82}$

$=1.51\times {10}^{-2}\mathrm{M}$

$\begin{array}{r}\left[{\mathrm{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{1.51\times {10}^{-2}}=6.6\times {10}^{-13}\mathrm{M}\end{array}$