Question
A solution that is 0.25 M in HN3 is also 0.55 M in NaN3. Given a pKa of 4.60 for HN3, what is the pH of this solution?
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Answer to a math question A solution that is 0.25 M in HN3 is also 0.55 M in NaN3. Given a pKa of 4.60 for HN3, what is the pH of this solution?
Rasheed
4.7109 Answers
To find the pH of the solution using the Henderson-Hasselbalch equation, we can substitute the given values into the equation:
\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
Given values:
-\text{pKa} = 4.60
-[\text{HA}] = 0.25 \, \text{M}
-[\text{A}^-] = 0.55 \, \text{M}
Substitute into Henderson-Hasselbalch equation:
\text{pH} = 4.60 + \log \left( \frac{0.55}{0.25} \right)
Calculate:
\text{pH} = 4.60 + \log(2.2) \approx 4.94
Therefore, the pH of the solution is approximately\boxed{4.94}
Given values:
-
-
-
Substitute into Henderson-Hasselbalch equation:
Calculate:
Therefore, the pH of the solution is approximately
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