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Sagot :
Sure, let's go through the solution step by step for determining the pH of the buffer after adding NaOH.
### Step 1: Determine Moles of Each Component
First, we need to calculate the moles of HOCl, NaOCl, and NaOH in the solution.
#### Moles of HOCl
The concentration of HOCl is 0.50 M and the volume is 100.0 mL.
[tex]\[ \text{Moles of HOCl} = \text{Concentration} \times \text{Volume} = 0.50 \, \text{M} \times 0.100 \, \text{L} = 0.050 \, \text{moles} \][/tex]
#### Moles of NaOCl
The concentration of NaOCl is 0.40 M and the volume is 100.0 mL.
[tex]\[ \text{Moles of NaOCl} = \text{Concentration} \times \text{Volume} = 0.40 \, \text{M} \times 0.100 \, \text{L} = 0.040 \, \text{moles} \][/tex]
#### Moles of NaOH
The concentration of NaOH is 1.0 M and the volume is 10.0 mL.
[tex]\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 1.0 \, \text{M} \times 0.010 \, \text{L} = 0.010 \, \text{moles} \][/tex]
### Step 2: Reaction and Adjust Moles
NaOH reacts quantitatively with HOCl in a 1:1 ratio:
[tex]\[ \text{NaOH} + \text{HOCl} \rightarrow \text{NaOCl} + \text{H}_2\text{O} \][/tex]
This reaction will reduce the moles of HOCl by the moles of NaOH added, and increase the moles of NaOCl correspondingly.
#### Moles of HOCl After Reaction
[tex]\[ \text{Moles of HOCl after reaction} = 0.050 \, \text{moles} - 0.010 \, \text{moles} = 0.040 \, \text{moles} \][/tex]
#### Moles of NaOCl After Reaction
[tex]\[ \text{Moles of NaOCl after reaction} = 0.040 \, \text{moles} + 0.010 \, \text{moles} = 0.050 \, \text{moles} \][/tex]
### Step 3: Calculate New Concentrations
Now, we need to calculate the new concentrations of HOCl and NaOCl after the reaction. The total volume of the solution is now:
[tex]\[ \text{Total Volume} = 100.0 \, \text{mL} + 10.0 \, \text{mL} = 110.0 \, \text{mL} = 0.110 \, \text{L} \][/tex]
#### Concentration of HOCl After Reaction
[tex]\[ \text{Concentration of HOCl} = \frac{\text{Moles of HOCl}}{\text{Total Volume}} = \frac{0.040 \, \text{moles}}{0.110 \, \text{L}} = 0.3636 \, \text{M} \][/tex]
#### Concentration of NaOCl After Reaction
[tex]\[ \text{Concentration of NaOCl} = \frac{\text{Moles of NaOCl}}{\text{Total Volume}} = \frac{0.050 \, \text{moles}}{0.110 \, \text{L}} = 0.4545 \, \text{M} \][/tex]
### Step 4: Use Henderson-Hasselbalch Equation
To find the pH, we use the Henderson-Hasselbalch equation:
[tex]\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \][/tex]
Given [tex]\( K_a \)[/tex] for HOCl is [tex]\( 3.5 \times 10^{-8} \)[/tex]:
[tex]\[ \text{p}K_a = -\log (3.5 \times 10^{-8}) \approx 7.46 \][/tex]
The ratio of the concentrations of NaOCl to HOCl (base to acid) is:
[tex]\[ \frac{[\text{Base}]}{[\text{Acid}]} = \frac{0.4545 \, \text{M}}{0.3636 \, \text{M}} \approx 1.25 \][/tex]
Therefore:
[tex]\[ \text{pH} = 7.46 + \log (1.25) = 7.46 + 0.0968 \approx 7.55 \][/tex]
So, the final pH of the buffer solution after adding 10.0 mL of 1.0 M NaOH is approximately 7.55.
### Step 1: Determine Moles of Each Component
First, we need to calculate the moles of HOCl, NaOCl, and NaOH in the solution.
#### Moles of HOCl
The concentration of HOCl is 0.50 M and the volume is 100.0 mL.
[tex]\[ \text{Moles of HOCl} = \text{Concentration} \times \text{Volume} = 0.50 \, \text{M} \times 0.100 \, \text{L} = 0.050 \, \text{moles} \][/tex]
#### Moles of NaOCl
The concentration of NaOCl is 0.40 M and the volume is 100.0 mL.
[tex]\[ \text{Moles of NaOCl} = \text{Concentration} \times \text{Volume} = 0.40 \, \text{M} \times 0.100 \, \text{L} = 0.040 \, \text{moles} \][/tex]
#### Moles of NaOH
The concentration of NaOH is 1.0 M and the volume is 10.0 mL.
[tex]\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 1.0 \, \text{M} \times 0.010 \, \text{L} = 0.010 \, \text{moles} \][/tex]
### Step 2: Reaction and Adjust Moles
NaOH reacts quantitatively with HOCl in a 1:1 ratio:
[tex]\[ \text{NaOH} + \text{HOCl} \rightarrow \text{NaOCl} + \text{H}_2\text{O} \][/tex]
This reaction will reduce the moles of HOCl by the moles of NaOH added, and increase the moles of NaOCl correspondingly.
#### Moles of HOCl After Reaction
[tex]\[ \text{Moles of HOCl after reaction} = 0.050 \, \text{moles} - 0.010 \, \text{moles} = 0.040 \, \text{moles} \][/tex]
#### Moles of NaOCl After Reaction
[tex]\[ \text{Moles of NaOCl after reaction} = 0.040 \, \text{moles} + 0.010 \, \text{moles} = 0.050 \, \text{moles} \][/tex]
### Step 3: Calculate New Concentrations
Now, we need to calculate the new concentrations of HOCl and NaOCl after the reaction. The total volume of the solution is now:
[tex]\[ \text{Total Volume} = 100.0 \, \text{mL} + 10.0 \, \text{mL} = 110.0 \, \text{mL} = 0.110 \, \text{L} \][/tex]
#### Concentration of HOCl After Reaction
[tex]\[ \text{Concentration of HOCl} = \frac{\text{Moles of HOCl}}{\text{Total Volume}} = \frac{0.040 \, \text{moles}}{0.110 \, \text{L}} = 0.3636 \, \text{M} \][/tex]
#### Concentration of NaOCl After Reaction
[tex]\[ \text{Concentration of NaOCl} = \frac{\text{Moles of NaOCl}}{\text{Total Volume}} = \frac{0.050 \, \text{moles}}{0.110 \, \text{L}} = 0.4545 \, \text{M} \][/tex]
### Step 4: Use Henderson-Hasselbalch Equation
To find the pH, we use the Henderson-Hasselbalch equation:
[tex]\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \][/tex]
Given [tex]\( K_a \)[/tex] for HOCl is [tex]\( 3.5 \times 10^{-8} \)[/tex]:
[tex]\[ \text{p}K_a = -\log (3.5 \times 10^{-8}) \approx 7.46 \][/tex]
The ratio of the concentrations of NaOCl to HOCl (base to acid) is:
[tex]\[ \frac{[\text{Base}]}{[\text{Acid}]} = \frac{0.4545 \, \text{M}}{0.3636 \, \text{M}} \approx 1.25 \][/tex]
Therefore:
[tex]\[ \text{pH} = 7.46 + \log (1.25) = 7.46 + 0.0968 \approx 7.55 \][/tex]
So, the final pH of the buffer solution after adding 10.0 mL of 1.0 M NaOH is approximately 7.55.
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