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Ch.17 - Applications of Aqueous Equilibria
Chapter 17, Problem 59

The pH of a solution of HN3 (Ka = 1.9 x10^-5) and NaN3 is 4.86. What is the molarity of NaN3 if the molarity of HN3 is 0.016 M?

Verified step by step guidance
1
Step 1: Recognize that the solution is a buffer solution, consisting of a weak acid (HN3) and its conjugate base (NaN3). Use the Henderson-Hasselbalch equation to find the relationship between pH, pKa, and the concentrations of the acid and its conjugate base.
Step 2: Calculate the pKa of HN3 using the given Ka value. The formula is \( \text{pKa} = -\log(\text{Ka}) \). Substitute \( \text{Ka} = 1.9 \times 10^{-5} \) into the formula to find pKa.
Step 3: Use the Henderson-Hasselbalch equation: \( \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \), where \([\text{A}^-]\) is the concentration of the conjugate base (NaN3) and \([\text{HA}]\) is the concentration of the weak acid (HN3).
Step 4: Substitute the known values into the Henderson-Hasselbalch equation: \( \text{pH} = 4.86 \), \( \text{pKa} \) from Step 2, and \([\text{HA}] = 0.016 \text{ M}\). Solve for \([\text{A}^-]\), the concentration of NaN3.
Step 5: Rearrange the equation to solve for \([\text{A}^-]\): \( \log\left(\frac{[\text{A}^-]}{0.016}\right) = \text{pH} - \text{pKa} \). Use the antilog to find \([\text{A}^-]\), which is the molarity of NaN3.