To solve the questions about the titration of HNO2 (a weak acid) with NaOH (a strong base), we will follow these steps:
(A) pH of the solution before titration
Initially, we have a 20.0 mL aliquot of 0.1 M HNO2. We can calculate the pH using the dissociation constant of HNO2.
The dissociation of HNO2 in water is represented by the equation:
\[ \text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^- \]
The acid dissociation constant (\(K_a\)) for HNO2 is approximately \(4.5 \times 10^{-4}\).
Using the expression for the acid dissociation constant:
\[ K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \]
Since the concentration of HNO2 initially is 0.1 M, and considering that \(x\) is the concentration of \(H^+\) that dissociates, we have:
\[ K_a = \frac{x^2}{(0.1 - x)} \]
Since \(K_a\) is small, we can assume \(x \ll 0.1\), which simplifies to:
\[ 4.5 \times 10^{-4} = \frac{x^2}{0.1} \]
Solving for \(x\):
\[ x^2 = 4.5 \times 10^{-5} \] \[ x = \sqrt{4.5 \times 10^{-5}} \approx 0.0067 \text{ M} \]
The pH is then:
\[ \text{pH} = -\log(0.0067) \approx 2.17 \]
(B) pH of the solution after 15.0 mL of NaOH has been added
After adding 15.0 mL of 0.1 M NaOH, we calculate the moles of HNO2 and NaOH:
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Moles of HNO2 initially: \[ Moles = 0.1, \text{M} \times 0.020, \text{L} = 0.002, \text{mol} \]
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Moles of NaOH added: \[ Moles = 0.1, \text{M} \times 0.015, \text{L} = 0.0015, \text{mol} \]
Now, we find the moles of HNO2 remaining and the moles of the conjugate base (NO2-) formed:
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Moles of HNO2 remaining: \[ 0.002, \text{mol} - 0.0015, \text{mol} = 0.0005, \text{mol} \]
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Moles of NO2- formed: \[ = 0.0015, \text{mol} \]
Now, we calculate the concentrations after mixing:
Total volume after adding NaOH: \(20, \text{mL} + 15, \text{mL} = 35, \text{mL} = 0.035, \text{L}\)
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Concentration of HNO2: \[ [\text{HNO}_2] = \frac{0.0005, \text{mol}}{0.035, \text{L}} \approx 0.0143, \text{M} \]
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Concentration of NO2-: \[ [\text{NO}_2^-] = \frac{0.0015, \text{mol}}{0.035, \text{L}} \approx 0.0429, \text{M} \]
We can now use the Henderson-Hasselbalch equation to find the pH:
\[ \text{pH} = pK_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) \]
First, calculate \(pK_a\):
\[ pK_a = -\log(4.5 \times 10^{-4}) \approx 3.35 \]
Now substitute into the equation:
\[ \text{pH} = 3.35 + \log\left(\frac{0.0429}{0.0143}\right) = 3.35 + \log(3) \approx 3.35 + 0.477 \approx 3.83 \]
(C) pH of the solution at equivalence point (after 20.0 mL of NaOH is added)
At the equivalence point, all the HNO2 has been neutralized:
- Moles of NaOH added: \[ = 0.1 , \text{M} \times 0.020 , \text{L} = 0.002, \text{mol} \]
At the equivalence point, we form 0.002 mol of NO2-. The total volume is now:
\[ 20, \text{mL} + 20, \text{mL} = 40, \text{mL} = 0.040, \text{L} \]
Concentration of NO2-:
\[ [\text{NO}_2^-] = \frac{0.002 \text{ mol}}{0.040 , \text{L}} = 0.050, \text{M} \]
NO2- acts as a weak base:
\[ \text{NO}_2^- + \text{H}_2\text{O} \rightleftharpoons \text{HNO}_2 + \text{OH}^- \]
To find the pOH, we need the base dissociation constant \(K_b\):
\[ K_w = K_a \times K_b \quad \text{(where $K_w = 1.0 \times 10^{-14}$)} \] \[ K_b = \frac{K_w}{K_a} \approx \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}} \approx 2.22 \times 10^{-11} \]
Using the expression for \(K_b\):
\[ K_b = \frac{[OH^-]^2}{[NO_2^-]} \Rightarrow [OH^-]^2 = K_b \times [NO_2^-] = 2.22 \times 10^{-11} \times 0.050 \] \[ [OH^-]^2 = 1.11 \times 10^{-12} \quad \Rightarrow \quad [OH^-] = \sqrt{1.11 \times 10^{-12}} \approx 1.05 \times 10^{-6}, \text{M} \]
Calculating the pOH:
\[ \text{pOH} = -\log(1.05 \times 10^{-6}) \approx 5.98 \]
Finally, since \(\text{pH} + \text{pOH} = 14\):
\[ \text{pH} = 14 - 5.98 \approx 8.02 \]
Summary of Results
(A) pH before titration: 2.17
(B) pH after adding 15.0 mL NaOH: 3.83
(C) pH at equivalence point: 8.02