Q.10.34 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ of each aqueous solution with the following $\left[{\mathrm{OH}}^{-}\right]$:

a. baking soda, $1.0\times {10}^{-6 }\mathrm{M}$

b. blood, $2.5\times {10}^{-7} \mathrm{M}$

c. milk, $4.0\times {10}^{-7} \mathrm{M}$

d. bleach, $2.1\times {10}^{-5} \mathrm{M}$

Step-by-Step Solution

Answer

a.  The [H₃O⁺] concentration is $1.0\times {10}^{-8 }\mathrm{M}$.

b. The [H₃O⁺] concentration is $0.4\times {10}^{-7 }\mathrm{M}$.

c. The [H₃O⁺] concentration is $0.25\times {10}^{-7} \mathrm{M}$

d. The [H₃O⁺] concentration is $0.476\times {10}^{-9} \mathrm{M}$.

Step 1: Part (a) Given information

The [OH^-] concentration is $1.0\times {10}^{-6 }\mathrm{M}$.

Step 2: Part (a) Explanation

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

The ionic product of water can be represented as follows:

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

Substituting the given values in the above equation.

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\times 1.0\times {10}^{-6}=1\times {10}^{-14} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{1\times {10}^{-14}}{1.0\times {10}^{-6}} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1\times {10}^{-8 }\mathrm{M} \end{gathered}$$

Thus, the [H₃O⁺] concentration is $1\times {10}^{-8} \mathrm{M}$.

Step 1: Part (b) Given information

The [OH^-] concentration of blood sample is $2.5\times {10}^{-7} \mathrm{M}$.

Step 2: Part (b) Explanation

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

The ionic product of water can be represented as follows:

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

Substituting the given value in the above equation.

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\times 2.5\times {10}^{-7}=1\times {10}^{-14} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{1\times {10}^{-14}}{2.5\times {10}^{-7}} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=0.4\times {10}^{-7 }\mathrm{M} \end{gathered}$$

Therefore,  the [H₃O⁺] concentration is $0.4\times {10}^{-7 }\mathrm{M}$.

Step 1: Part (c) Given information

The [OH^-] concentration is $4.0\times {10}^{-7 }\mathrm{M}$.

Step 2: Part (c) Explanation

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

The ionic product of water can be represented as follows:

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

Substituting the given value in the above equation.

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\times 4.0\times {10}^{-7}=1\times {10}^{-14} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{1\times {10}^{-14}}{4.0\times {10}^{-7}} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=0.25\times {10}^{-7 }\mathrm{M} \end{gathered}$$

Therefore,  the [H₃O⁺] concentration is $0.25\times {10}^{-7} \mathrm{M}$.

Step 1: Part (d) Given information

The [OH^-] concentration is $2.1\times {10}^{-5} \mathrm{M}$.

Step 2: Part (d) Explanation

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

The ionic product of water can be represented as follows:

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

Substituting the given value in the above equation.

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\times 2.1\times {10}^{-5}=1\times {10}^{-14} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{1\times {10}^{-14}}{2.1\times {10}^{-5}} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=0.476\times {10}^{-9 }\mathrm{M} \end{gathered}$$

Thus,  the [H₃O⁺] concentration is $0.476\times {10}^{-9} \mathrm{M}$.