Q.10.91 - An Introduction to General, Organic, and Biological Chemistry

Question

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

a. 3.40

b. 6.00

c. 8.0

d. 11.0

e. 9.20

Step-by-Step Solution

Answer

Part a.$\begin{array}{r}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=3.98\times {10}^{-4},\left[{\mathrm{OH}}^{-}\right]=2.51\times {10}^{-11}\end{array}$

Part b. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=6.02\times {10}^{-7},\left[{\mathrm{OH}}^{-}\right]=1.66\times {10}^{-8}$

Part c. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-8},\left[{\mathrm{OH}}^{-}\right]={10}^{-6}$

Part d. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-11},\left[{\mathrm{OH}}^{-}\right]={10}^{-3}$

Part e.$\begin{array}{r}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=6.30\times {10}^{-10},\left[{\mathrm{OH}}^{-}\right]=1.58\times {10}^{-5}\end{array}$

Step 1: Given Information (Part a)

The representation of a chemical reaction in the type of substance is known as a chemical equation. The equation in which the quantity of particles of the relative multitude of atoms is equivalent on the two sides of the equation is known as a reasonable chemical equation.

Step 2: Explanation Part (a)

Given pH = $3.4$

We know,

$pH=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$=$\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=-3.4$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=3.98\times {10}^{-4}$

Now using,

$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right]=\frac{1\times {10}^{-14}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{1\times {10}^{-14}}{3.98\times {10}^{-4}} \\ \left[{\mathrm{OH}}^{-}\right]=2.5\times {10}^{-11} \end{gathered}$$

Step 3: Explanation Part (b)

Given pH = $6$

We know,

$pH=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=-6$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=6.02\times {10}^{-7}$

Now using,

$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right]=\frac{1\times {10}^{-14}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{1\times {10}^{-14}}{6.02\times {10}^{-7}} \\ \left[{\mathrm{OH}}^{-}\right]=1.66\times {10}^{-8} \end{gathered}$$

Step 4: Explanation Part (c)

Given pH = $8$

We know,

$pH=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=-8$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-8}$

Using,

$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right]=\frac{{10}^{-14}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{{10}^{-14}}{{10}^{-8}} \\ \left[{\mathrm{OH}}^{-}\right]={10}^{-6} \end{gathered}$$

Step 5: Explanation Part (d)

Given pH = $11$

We know,

$\text{ pH}-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=-11$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-11}$

Using,

$\left[{\mathrm{OH}}^{-}\right]=\frac{{10}^{-14}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{{10}^{-14}}{{10}^{-11}}={10}^{-3}$

Step 6: Explanation Part (e)

Given pH = $9.2$

We know,

$\text{ pH=}-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=-9.2$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=6.3\times {10}^{-10}$

Using,

$$\begin{gathered} \begin{array}{r}\left[{\mathrm{OH}}^{-}\right]=\frac{1\times {10}^{-14}}{{\mathrm{H}}_3{\mathrm{O}}^{+}}\end{array} \\ \left[{\mathrm{OH}}^{-}\right]=\frac{{10}^{-14}}{6.3\times {10}^{-10}}=2.51\times {10}^{-11} \end{gathered}$$