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To determine the pH of a 0.320 M solution of calcium nitrite, [tex]\( \text{Ca(NO}_2\text{)}_2 \)[/tex], with a given [tex]\( K_a \)[/tex] for nitrous acid ([tex]\( \text{HNO}_2 \)[/tex]) of [tex]\( 4.5 \times 10^{-4} \)[/tex]:
### Step-by-Step Solution:
1. Write the Dissociation Reaction:
Calcium nitrite dissociates completely in water:
[tex]\[ \text{Ca(NO}_2\text{)}_2 \rightarrow \text{Ca}^{2+} + 2\text{NO}_2^- \][/tex]
2. Determine the Concentration of [tex]\( \text{NO}_2^- \)[/tex]:
Since each formula unit of [tex]\( \text{Ca(NO}_2\text{)}_2 \)[/tex] produces two nitrite ions ([tex]\( \text{NO}_2^- \)[/tex]):
[tex]\[ \text{Concentration of } \text{NO}_2^- = 2 \times 0.320\, \text{M} = 0.640\, \text{M} \][/tex]
3. Set Up the Equilibrium:
Nitrite ions ([tex]\( \text{NO}_2^- \)[/tex]) can react with water to form nitrous acid ([tex]\( \text{HNO}_2 \)[/tex]) and hydroxide ions ([tex]\( \text{OH}^- \)[/tex]). However, an easier approach is to determine the hydrogen ion concentration ([tex]\( \text{H}^+ \)[/tex]) using the [tex]\( K_a \)[/tex] value of nitrous acid ([tex]\( \text{HNO}_2 \)[/tex]) and the concentration of [tex]\(\text{NO}_2^- \)[/tex].
The dissociation of [tex]\( \text{HNO}_2 \)[/tex] is:
[tex]\[ \text{HNO}_2 \leftrightarrow \text{H}^+ + \text{NO}_2^- \][/tex]
The equilibrium expression for [tex]\( K_a \)[/tex] is:
[tex]\[ K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \][/tex]
4. Assumption for Simplification:
Assuming the reaction [tex]\( \text{NO}_2^- \leftrightarrow \text{HNO}_2 + \text{OH}^- \)[/tex] reaches an equilibrium where [tex]\( [\text{H}^+] \approx [\text{OH}^-] \)[/tex].
5. Calculate Hydrogen Ion Concentration:
To find [tex]\( \text{H}^+ \)[/tex] ions, we assume that at equilibrium, [tex]\( [\text{H}^+] \)[/tex] from the dissociation is in square root relation of the product of [tex]\( K_a \)[/tex] and the nitrite ion concentration:
[tex]\[ [\text{H}^+] = \sqrt{K_a \times [\text{NO}_2^-]} = \sqrt{4.5 \times 10^{-4} \times 0.640} \approx 0.01697056 \, \text{M} \][/tex]
6. Calculate the pH:
The pH is calculated using the formula:
[tex]\[ \text{pH} = -\log [\text{H}^+] \][/tex]
Substituting the concentration of [tex]\( \text{H}^+ \)[/tex]:
[tex]\[ \text{pH} = -\log (0.01697056) \approx 1.77 \][/tex]
### Conclusion:
Thus, the pH of a 0.320 M solution of [tex]\( \text{Ca(NO}_2\text{)}_2 \)[/tex] is approximately 1.77.
### Step-by-Step Solution:
1. Write the Dissociation Reaction:
Calcium nitrite dissociates completely in water:
[tex]\[ \text{Ca(NO}_2\text{)}_2 \rightarrow \text{Ca}^{2+} + 2\text{NO}_2^- \][/tex]
2. Determine the Concentration of [tex]\( \text{NO}_2^- \)[/tex]:
Since each formula unit of [tex]\( \text{Ca(NO}_2\text{)}_2 \)[/tex] produces two nitrite ions ([tex]\( \text{NO}_2^- \)[/tex]):
[tex]\[ \text{Concentration of } \text{NO}_2^- = 2 \times 0.320\, \text{M} = 0.640\, \text{M} \][/tex]
3. Set Up the Equilibrium:
Nitrite ions ([tex]\( \text{NO}_2^- \)[/tex]) can react with water to form nitrous acid ([tex]\( \text{HNO}_2 \)[/tex]) and hydroxide ions ([tex]\( \text{OH}^- \)[/tex]). However, an easier approach is to determine the hydrogen ion concentration ([tex]\( \text{H}^+ \)[/tex]) using the [tex]\( K_a \)[/tex] value of nitrous acid ([tex]\( \text{HNO}_2 \)[/tex]) and the concentration of [tex]\(\text{NO}_2^- \)[/tex].
The dissociation of [tex]\( \text{HNO}_2 \)[/tex] is:
[tex]\[ \text{HNO}_2 \leftrightarrow \text{H}^+ + \text{NO}_2^- \][/tex]
The equilibrium expression for [tex]\( K_a \)[/tex] is:
[tex]\[ K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \][/tex]
4. Assumption for Simplification:
Assuming the reaction [tex]\( \text{NO}_2^- \leftrightarrow \text{HNO}_2 + \text{OH}^- \)[/tex] reaches an equilibrium where [tex]\( [\text{H}^+] \approx [\text{OH}^-] \)[/tex].
5. Calculate Hydrogen Ion Concentration:
To find [tex]\( \text{H}^+ \)[/tex] ions, we assume that at equilibrium, [tex]\( [\text{H}^+] \)[/tex] from the dissociation is in square root relation of the product of [tex]\( K_a \)[/tex] and the nitrite ion concentration:
[tex]\[ [\text{H}^+] = \sqrt{K_a \times [\text{NO}_2^-]} = \sqrt{4.5 \times 10^{-4} \times 0.640} \approx 0.01697056 \, \text{M} \][/tex]
6. Calculate the pH:
The pH is calculated using the formula:
[tex]\[ \text{pH} = -\log [\text{H}^+] \][/tex]
Substituting the concentration of [tex]\( \text{H}^+ \)[/tex]:
[tex]\[ \text{pH} = -\log (0.01697056) \approx 1.77 \][/tex]
### Conclusion:
Thus, the pH of a 0.320 M solution of [tex]\( \text{Ca(NO}_2\text{)}_2 \)[/tex] is approximately 1.77.
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