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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?

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Rasheed

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110 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}

(concentration of the weak acid)
-

[\text{A}^-] = 0.55 \, \text{M}

(concentration of the conjugate base)

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}

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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?

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?

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