Q.10.87 - An Introduction to General, Organic, and Biological Chemistry

Question

Determine the pH for the following solutions:

a. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0\times {10}^{-11}\mathrm{M}$

b. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=5.0\times {10}^{-2}\mathrm{M}$

c. $\left[{\mathrm{OH}}^{-}\right]=3.5\times {10}^{-4}\mathrm{M}$

d. $\left[{\mathrm{OH}}^{-}\right]=0.005\mathrm{M}$

Step-by-Step Solution

Answer

Part a.$11$

Part b.$1.3$

Part c.$10.5$

Part d.$11.7$

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 multitude of atoms is equivalent on the two sides of the equation is known as a fair chemical equation.

Step 2: Explanation Part (a)

We know,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0\times {10}^{-11}\mathrm{M}$

Using,

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

$$\begin{gathered} \mathrm{pH}=-\log \left(1.0\times {10}^{-11}\right) \\ =11 \end{gathered}$$

Step 3: Explanation Part (b)

We know,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=5.0\times {10}^{-2}\mathrm{M}$

Using,

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

$$\begin{gathered} \mathrm{pH}=-\log \left(5.0\times {10}^{-2}\right) \\ =1.30 \end{gathered}$$

Step 4: Explanation Part (c)

We know,

$\left[{\mathrm{OH}}^{-}\right]=3.5\times {10}^{-4}\mathrm{M}$

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

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{1\times {10}^{-14}}{\left[{\mathrm{OH}}^{-}\right]}=\frac{1\times {10}^{-14}}{3.5\times {10}^{-4}}=2.85\times {10}^{-11}$

Using,

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ pH=-\log (2.85\times {10}^{-11}) \\ pH=10.5 \end{gathered}$$

Step 5: Explanation Part (d)

We know,

$\left[{\mathrm{OH}}^{-}\right]=0.005\mathrm{M}$

Calculating $H_3{O}^{+}$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{1\times {10}^{-14}}{\left[{\mathrm{OH}}^{-}\right]}=\frac{1\times {10}^{-14}}{0.005}=2\times {10}^{-12}$

Using,

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

$$\begin{gathered} \mathrm{pH}=-\log \left(2.0\times {10}^{-12}\right) \\ pH=11.7 \end{gathered}$$