To calculate the pH of the resulting solution after mixing \(16.0 , \text{mL}\) of \(0.160 , \text{M HCl}\) with \(26.0 , \text{mL}\) of \(0.210 , \text{M NOH}\), we need to follow these steps:
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Calculate the moles of HCl and NOH:
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For HCl: \[ \text{moles of HCl} = \text{Volume in L} \times \text{Concentration} = 0.016 , \text{L} \times 0.160 , \text{mol/L} = 0.00256 , \text{mol} \]
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For NOH: \[ \text{moles of NOH} = \text{Volume in L} \times \text{Concentration} = 0.026 , \text{L} \times 0.210 , \text{mol/L} = 0.00546 , \text{mol} \]
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Determine the reaction between HCl and NOH:
HCl is a strong acid and NOH is a strong base. The neutralization reaction is: \[ \text{HCl} + \text{NOH} \rightarrow \text{H}_2\text{O} + \text{NOCl} \]
- Find the limiting reagent:
Since we have \(0.00256 , \text{mol HCl}\) and \(0.00546 , \text{mol NOH}\), HCl is the limiting reagent.
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Calculate the moles of acid and base remaining:
Moles of NOH after reaction: \[ \text{Remaining NOH} = 0.00546 , \text{mol} - 0.00256 , \text{mol} = 0.00290 , \text{mol} \]
Moles of HCl after reaction: \[ \text{Remaining HCl} = 0, \text{mol} , \text{(since it's the limiting reagent)} \]
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Calculate the concentration of NOH remaining:
Total volume of the resulting solution: \[ \text{Total volume} = 16.0 , \text{mL} + 26.0 , \text{mL} = 42.0 , \text{mL} = 0.042 , \text{L} \]
Concentration of remaining NOH: \[ \text{Concentration of NOH} = \frac{0.00290 , \text{mol}}{0.042 , \text{L}} = 0.06905 , \text{M} \]
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Calculate the pOH of the solution:
\[ \text{pOH} = -\log[\text{NOH}] = -\log[0.06905] \approx 1.16 \]
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Calculate the pH of the solution:
Since \( \text{pH} + \text{pOH} = 14 \): \[ \text{pH} = 14 - 1.16 \approx 12.84 \]
Final Answer: The pH of the resulting solution is approximately 12.84.