To calculate the pH of the resulting solution when 16.0 mL of 0.160 M HCl is mixed with 21.0 mL of 0.160 M NaOH, we can follow these steps:
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Calculate the moles of HCl and NaOH:
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Moles of HCl: \[ \text{Moles of HCl} = \text{Volume (L)} \times \text{Concentration (M)} = 0.016 , \text{L} \times 0.160 , \text{M} = 0.00256 , \text{moles} \]
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Moles of NaOH: \[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.021 , \text{L} \times 0.160 , \text{M} = 0.00336 , \text{moles} \]
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Determine which reactant is in excess:
- The balanced reaction for HCl and NaOH is: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
Since both HCl and NaOH react in a 1:1 molar ratio, we can compare their moles:
- \( \text{Moles of NaOH} = 0.00336 \)
- \( \text{Moles of HCl} = 0.00256 \)
Since \( 0.00336 > 0.00256 \), NaOH is in excess.
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Calculate the moles of excess NaOH: \[ \text{Excess NaOH} = \text{Moles of NaOH} - \text{Moles of HCl} = 0.00336 - 0.00256 = 0.00080 , \text{moles of NaOH} \]
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Calculate the total volume of the solution: \[ \text{Total Volume} = 16.0 , \text{mL} + 21.0 , \text{mL} = 37.0 , \text{mL} = 0.037 , \text{L} \]
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Calculate the concentration of excess NaOH: \[ \text{Concentration of excess NaOH} = \frac{\text{Moles of excess NaOH}}{\text{Total Volume (L)}} = \frac{0.00080 , \text{moles}}{0.037 , \text{L}} \approx 0.0216 , \text{M} \]
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Calculate the pOH and then the pH: \[ \text{pOH} = -\log[\text{OH}^-] = -\log(0.0216) \approx 1.666 \] \[ \text{pH} = 14 - \text{pOH} \approx 14 - 1.666 \approx 12.334 \]
Thus, the pH of the resulting solution is approximately 12.33.