Q. 10.40 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the $\mathrm{pH}$ of each solution given the following:

a. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1\times {10}^{-8}\mathrm{M}$

b. $\left|{\mathrm{H}}_3{\mathrm{O}}^{+}\right|=5\times {10}^{-6}\mathrm{M}$

c. $\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-2}\mathrm{M}$

d. $\left[{\mathrm{OH}}^{-}\right]=8.0\times {10}^{-1}\mathrm{M}$

e.$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=4.7\times {10}^{-2}\mathrm{M}$

f. $\left[{\mathrm{OH}}^{-}\right]=3.9\times {10}^{-6}\mathrm{M}$

Step-by-Step Solution

Answer

a) The pH value of the solution is determined as $8$

b)The pH value of the solution is determined as $5.3011$

c) The pH value of the solution is determined as $2$

d) The pH value of the solution is determined as $2.097$

e) The pH value of the solution is determined as $1.328$

f) The pH value of the solution is determined as $5.409$

Step 1 : Introduction (part a)

The objective is to determine the pH values of the solution

Step 2 : Explanation (part a)

The pH scale is a logarithmic scale that measures the acidity of water. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$of aqueous solutions. Mathematically expression of $\mathrm{pH}$ is the negative logarithmic (base 10 ) of the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

$$\begin{gathered} pH=-\log \left[H_3{O}^{+}\right] \\ pOH=-\log \left[O{H}^{-}\right] \\ pH+pOH =14 \\ pH=14-pOH \end{gathered}$$

Acidic solution $=\mathrm{pH}<7.0\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]>1.0\times {10}^{-7}\mathrm{M}$

Neutral solution $=\mathrm{pH}=7.0\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0\times {10}^{-7}\mathrm{M}$

Basic solution $=\mathrm{pH}>7.0\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]<1.0\times {10}^{-7}\mathrm{M}$

(a)

The given concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$is $1.0\times {10}^{-8}\mathrm{M}$.

$$\begin{gathered} [H_3 O^{+}]=1.0 {10}^{-8} M \\ pH=-\log \left[H_3{O}^{+}\right] \\ =-\log (1.0 \times {10}^{-8}) \end{gathered}$$

Step 3 : Introduction (part b)

The objective is to determine the pH values of the solution 

Step 4 : Explanation (part b)

(b)

The given concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$is $5\times {10}^{-6}\mathrm{M}$

$$\begin{gathered} [H_3 O^{+}]=5 \times {10}^{-6}M \\ pH=-\log [H_3 O^{+}] \\ =-\log (5 \times {10}^{-6}) \\ pH=6-0.6989 \\ =5.3011 \end{gathered}$$

Therefore, $\mathrm{pH}=5.3011$

Step 5 : Introduction (part c)

The objective is to determine the pH values of the solution 

Step 6 : Explanation (part c)

(c)

The given concentration of ${\mathrm{OH}}^{-}$is $1\times {10}^{-2}\mathrm{M}$

$$\begin{gathered} pOH=-\log \left[O{H}^{-}\right] \\ =-\log (1 \times {10}^{-2}) \\ pOH=2 \end{gathered}$$

Therefore, $\mathrm{pOH}=2$

Step 7 : Introduction (part d)

The objective is to determine the pH values of the solution 

Step 8 : Explanation (part d)

(d)

The given concentration of ${\mathrm{OH}}^{-}$is $8.0\times {10}^{-3}\mathrm{M}$

$$\begin{gathered} pOH =-\log \left[O{H}^{-}\right] \\ =-\log (8.0 \times {10}^{-3}) \\ pOH =3-0.9030 \\ =2.097 \end{gathered}$$

Therefore, $\mathrm{pOH}=2.097$

Step 9: Introduction (part e)

The objective is to determine the pH values of the solution 

Step 10 : Explanation (part e)

(e)

The given concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$is $4.7\times {10}^{-2}\mathrm{M}$.

$$\begin{gathered} pH=-\log [H_3 O^{+}] \\ =-\log (4.7 \times {10}^{-2}) \\ pH=2-0.672 \\ =1.328 \end{gathered}$$

As a result, $\mathrm{pH}=1.328$

Step 11 : Introduction (part f)

The objective is to determine the pH values of the solution 

Step 12 : Explanation (part f)

(f)

The given concentration of ${\mathrm{OH}}^{-}$is $3.9\times {10}^{-6}\mathrm{M}$

$$\begin{gathered} pOH=-\log \left[O{H}^{-}\right] \\ =-\log (3.9 \times {10}^{-6}) \\ pOH=6-0.5910 \\ =5.409 \end{gathered}$$

As a result, $\mathrm{pOH}=5.409$.