The pH of a solution of HN3 (ka=1.9*10^-5) and NaN3 is 4.86.
What is the molarity of NAN3 if the molarity of HN3 is 0.020M?
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The Correct Answer and Explanation is :
To solve this problem, we use the Henderson-Hasselbalch equation for buffer solutions:
[
\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
]
Where:
- (\text{pH} = 4.86) (given)
- (\text{pKa} = -\log K_a = -\log(1.9 \times 10^{-5}))
- ([\text{HA}] = [\text{HN}_3] = 0.020\, \text{M}) (given)
- ([\text{A}^-] = [\text{NaN}_3]) (unknown)
Step 1: Calculate (\text{pKa})
[
\text{pKa} = -\log(1.9 \times 10^{-5}) = 4.72
]
Step 2: Substitute values into the Henderson-Hasselbalch equation
[
4.86 = 4.72 + \log \left( \frac{[\text{NaN}_3]}{0.020} \right)
]
Step 3: Solve for (\frac{[\text{NaN}_3]}{0.020})
[
4.86 – 4.72 = \log \left( \frac{[\text{NaN}_3]}{0.020} \right)
]
[
0.14 = \log \left( \frac{[\text{NaN}_3]}{0.020} \right)
]
Step 4: Remove the logarithm
[
10^{0.14} = \frac{[\text{NaN}_3]}{0.020}
]
[
1.38 = \frac{[\text{NaN}_3]}{0.020}
]
Step 5: Solve for ([\text{NaN}_3])
[
[\text{NaN}_3] = 1.38 \times 0.020 = 0.0276\, \text{M}
]
Final Answer:
The molarity of (\text{NaN}_3) is:
[
\boxed{0.0276 \, \text{M}}
]
Explanation (300 words)
The problem involves a buffer solution containing a weak acid (HN(_3)) and its conjugate base (NaN(_3)). Such solutions resist changes in pH, and the Henderson-Hasselbalch equation is used to relate the pH, the acid dissociation constant ((K_a)), and the concentrations of the acid ((HA)) and its conjugate base ((A^-)).
First, the (K_a) of HN(_3) is used to calculate the pKa ((pKa = -\log K_a)), which measures the strength of the weak acid. Knowing the pH (4.86) and the concentration of HN(_3) (0.020 M), we rearranged the Henderson-Hasselbalch equation to solve for the ratio of the base (([\text{NaN}_3])) to the acid (([\text{HN}_3])).
After isolating the ratio, the logarithmic function was converted to exponential form, allowing us to directly solve for ([\text{NaN}_3]). The result, 0.0276 M, indicates the concentration of NaN(_3) required to maintain the given pH of the solution.
This calculation demonstrates the importance of the conjugate base in stabilizing the solution’s pH. If the concentration of NaN(_3) changes, the buffer’s effectiveness at maintaining pH decreases.