Question

Ammonia is a weak base with a ${\mathbf{K}}_b$ of .$\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{-5}$ A 140mL sample of a 0.175 M solution of aqueous ammonia is titrated with a 0.106 M solution of the strong acid HCl. The reaction is

${\mathbf{NH}}_3\left(\mathbf{aq}\right)\mathbf{+}\mathbf{HCl}\left(\mathbf{aq}\right)\to {\mathbf{NH}}_4^{+}\left(\mathbf{aq}\right)\mathbf{+}{\mathbf{Cl}}^{-}\left(\mathbf{aq}\right)$

Compute the pH of the titration solution before any acid is added, when the titration is at the half-equivalence point, when the titration is at the equivalence point, and when the titration is 1.00 mL past the equivalence point

Short answer

The pH of the solution after the addition of base is 11.25.

Step-by-step solution

Definition of pH

• The “negative logarithm” of hydrogen ion$\left({\mathbf{H}}^{+}\right)$concentration is termed as pH.

Mathematically,$\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]$

• The negative logarithm of the hydroxide$\left({\mathbf{OH}}^{-}\right)$ion concentration of a solution is termed as pOH.

Mathematically, $\text{pOH}=-\log \left[{\text{OH}}^{\text{-}}\right]$

Construction of ICE table 

The complete equation is:

${\text{NH}}_{\text{3}}\left(\text{aq}\right)\text{+HCl}\left(\text{aq}\right)\to {\text{NH}}_{\text{4}}^{\text{+}}\left(\text{aq}\right){\text{+Cl}}^{\text{-}}\left(\text{aq}\right)$

Since${\text{NH}}_3$ is a weak base, let the change in equilibrium concentration of base be y mol/L. The value of $K_b$is .$1.8\times {10}^{-5}$

The initial, change and equilibrium (ICE) table

Substitute the following values for the equilibrium constant in the equation:

$\begin{array}{c}{\text{K}}_{\text{b}}\text{=}\frac{\left[{\text{NH}}_{\text{4}}^{\text{-}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{NH}}_{\text{3}}\right]}\\ {\text{1.8×10}}^{\text{-5}}\text{=}\frac{{\text{y}}^{\text{2}}}{\left(\text{0.175-y}\right)}\mathrm{..........}\text{(i)}\end{array}$

Since ${\mathrm{K}}_{\mathrm{b}}$ is a small value, we can assume that data-custom-editor="chemistry" $y<<0.175$. Thus, we can ignoredata-custom-editor="chemistry" $y$ in the denominator and modify the equation.

Calculation of pOH

Equation is further simplified.

$\begin{array}{c}1.8\times {10}^{-5}=\frac{y^2}{0.175-y}\\ y=1.77\times {10}^{-3}\\ \left[{\text{OH}}^{\text{-}}\right]=1.77\times {10}^{-3}\end{array}$

Let us calculate the $\text{pOH}$ of the solution by applying the following formula

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{\text{-}}\right]\\ =-\log [1.77\times {10}^{-3}]\\ \text{pOH}=2.752\end{array}$

Calculation of pH

Now calculate the pH by applying the formula below.

$\begin{array}{c}14=\text{pH}+\text{pOH}\\ \text{pH}=14-\text{pOH}\\ =14-2.752\\ =11.25\end{array}$

Hence, the pH of the solution after the addition of the base is $11.25$.