Q.10.88 - An Introduction to General, Organic, and Biological Chemistry

Question

Determine the pH for the following solutions:

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

b.$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=4.2\times {10}^{-3}\mathrm{M}$

c.$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=0.0001\mathrm{M}$

d.$\left[{\mathrm{OH}}^{-}\right]=8.5\times {10}^{-9}\mathrm{M}$

Step-by-Step Solution

Answer

Part a. $7$

Part b. $2.38$

Part c. $4$

Part d. $5.93$

Step 1: Given Information (Part a)

pH or potential of hydrogen is a scale used to specify the acidity or basicity of a fluid solution. Acidic solutions are estimated to have lower pH values than basic or alkaline solutions

Step 2: Explanation Part (a)

We know,

$\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-7}\mathrm{M}$

Calculating $H_3{O}^{+}$,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{{\displaystyle 1\times {10}^{-14}}}{\left[{\mathrm{OH}}^{-}\right]}=\frac{1\times {10}^{-14}}{1\times {10}^{-7}}=1\times {10}^{-7}M$

Using,

$\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

$$\begin{gathered} \mathrm{pH}=-\log \left(1\times {10}^{-7}\right) \\ pH=7 \end{gathered}$$

Step 3: Explanation Part (b)

We know,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=4.2\times {10}^{-3}\mathrm{M}$

Using,

$\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

$$\begin{gathered} \mathrm{pH}=-\log \left(4.2\times {10}^{-3}\right) \\ pH=2.38 \end{gathered}$$

Step 4: Explanation Part (c)

We know,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=0.0001\mathrm{M}$

Using,

$\mathrm{pH}=-log\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

$$\begin{gathered} \mathrm{pH}=-\log \left(0.0001\right) \\ pH=4 \end{gathered}$$

Step 5: Explanation Part (d)

We know,

$O{H}^{-}=8.5\times {10}^{-9}$

Calculating $H_3{O}^{+}$,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{1\times {10}^{-14}}{\left[{\mathrm{OH}}^{-}\right]}=\frac{1\times {10}^{-14}}{8.5\times {10}^{-9}}=1.17\times {10}^{-6}\mathrm{M}$

Using,

$\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

$$\begin{gathered} \mathrm{pH}=-\log \left(1.17\times {10}^{-6}\right) \\ pH=5.93 \end{gathered}$$