Question

Consider the titration of 100.0 mL of 0.100 M H₂NNH₂ (Kb = 3.0 x 10^-6) with 0.200 M HNO₃. Calculate the pH of the resulting solution after each of the following volumes of HNO₃ has been added.
a. 0.0 mL d. 40.0 mL
b. 20.0 mL e. 50.0 mL
c. 25.0 mL f. 100.0 mL

Short answer

The pH of the resulting solution can be calculated.

Step-by-step solution

The pH of the resulting solution after 0 mL volumes of HNO3 has been added

The equation can be denoted as:

$H_{2}NN{H}_2+H_{2}O\rightleftharpoons H_{2}NN{H_3}^{+}+O{H}^{-}$

It can be denoted as:

The pH of the resulting solution after 20 mL volumes of HNO3 has been added 

The equation can be denoted as:

$H_{2}NN{H}_2+H_{2}O\rightleftharpoons H_{2}NN{H_3}^{+}+O{H}^{-}$

The mole number of hydrazine present can be $\frac{0.1 mol}{1000 mL}=100 mL=0.01 mol.$Themole number of H⁺ added can be $\frac{0.2 mol}{1000 mL}\times 20 mL=0.004 mol.$The number of moles of left hydrazine can be $0.01-0.004=0.006 mol.$The mole number of conjugate acid formation can be 0.004 mol.

$$\begin{gathered} K_a=\frac{K_w}{K_b} \\ =\frac{1\times {10}^{-14}}{3\times {10}^{-6}} \\ {K}_a=3.3\times {10}^{-9} \\ pH=p{K}_a+\log \frac{\left[H_{2}NN{H}_2\right]}{\left[H_{2}NN{H_3}^{+}\right]} \\ =-\log (3.3\times {10}^{-9})+\log \left(\frac{0.006/V}{0.004/V}\right) \\ =8.48+0.18 \\ pH=8.66 \end{gathered}$$

The pH of the resulting solution after 25 mL volumes of HNO3 has been added

The equation can be denoted as:

$H_{2}NN{H}_2+H_{2}O\rightleftharpoons H_{2}NN{H_3}^{+}+O{H}^{-}$

The mole number of hydrazine present can be $\frac{0.1 mol}{1000 mL}\times 100 mL=0.01 mol.$The mole number of H⁺ added can be $\frac{0.2 mol}{1000 mL}\times 25 mL=0.005 mol.$The number of moles of left hydrazine can be $0.01-0.005=0.005 mol.$The mole number of conjugate acid formation can be 0.005 mol.

$$\begin{gathered} K_a=\frac{K_w}{K_b} \\ =\frac{1\times {10}^{-14}}{3\times {10}^{-6}} \\ {K}_a=3.3\times {10}^{-9} \\ pH=p{K}_a+\log \frac{\left[H_{2}NN{H}_2\right]}{\left[H_{2}NN{H_3}^{+}\right]} \\ =-\log (3.3\times {10}^{-9})+\log \left(\frac{0.005/V}{0.005/V}\right) \\ pH=8.48 \end{gathered}$$

The pH of the resulting solution after 40 mL volumes of HNO3 has been added 

The equation can be denoted as:

$H_{2}NN{H}_2+H_{2}O\rightleftharpoons H_{2}NN{H_3}^{+}+O{H}^{-}$

The mole number of hydrazine present can be $\frac{0.1 mol}{1000 mL}=100 mL=0.01 mol.$The mole number of H⁺ added can be $\frac{0.2 mol}{1000 mL}\times 40 mL=0.008 mol.$The number of moles of left hydrazine can be $0.01-0.008=0.002 mol.$The mole number of conjugate acid formation can be 0.008 mol.

$$\begin{gathered} K_a=\frac{K_w}{K_b} \\ =\frac{1\times {10}^{-14}}{3\times {10}^{-6}} \\ {K}_a=3.3\times {10}^{-9} \\ pH=p{K}_a+\log \frac{\left[H_{2}NN{H}_2\right]}{\left[H_{2}NN{H_3}^{+}\right]} \\ =-\log (3.3\times {10}^{-9})+\log \left(\frac{0.002/V}{0.008/V}\right) \\ =8.48+(-0.6) \\ pH=7.88 \end{gathered}$$

The pH of the resulting solution after 50 mL volumes of HNO3 has been added

The equation can be denoted as:

$H_{2}NN{H}_2+H_{2}O\rightleftharpoons H_{2}NN{H_3}^{+}+O{H}^{-}$

The mole number of hydrazine present can be $\frac{0.1 mol}{1000 mL}=100 mL=0.01 mol.$The mole number of H⁺ added can be $\frac{0.2 mol}{1000 mL}\times 50 mL=0.01 mol.$The number of moles of left hydrazine can be $0.01-0.01=0.0 mol.$The mole number of conjugate acid formation can be 0.01 mol.

$$\begin{gathered} K_b=\frac{\left[H^{+}\right]\left[H_{2}NN{H}_2\right]}{\left[H_{2}NN{H_3}^{+}\right]} \\ 3.3\times {10}^{-9}=\frac{x^2}{(0.0667-x)}\approx \frac{x^2}{0.0667} \\ x=\left[H^{+}\right]=1.5\times {10}^{-5}M \\ pOH=-\log \left[H^{+}\right] \\ =-\log (1.5\times {10}^{-5}) \\ pH=4.82 \end{gathered}$$

The pH of the resulting solution after 100 mL volumes of HNO3 has been added

The mole number of H⁺ added can be $\frac{0.2 mol}{1000 mL}\times 100 mL=0.02 mol.$The mole number of conjugate acid formation can be 0.01 mol. The number of excess H⁺ moles can be $0.02-0.01=0.01 mol.$

The H⁺ amount added from weak acid can be neglected due to its low dissociation constant. It happens only when weak and strong acids are present in a solution.

$$\begin{gathered} {\left[H^{+}\right]}_{excess}=\frac{0.01 mol}{(100+100)mL} \\ {\left[H^{+}\right]}_{excess}=0.05 M \\ pH=-\log \left[H^{+}\right] \\ =-\log (0.05) \\ pH=1.3 \end{gathered}$$