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What is the pH of a solution made by mixing 50.0 mL of a 0.20 M solution of HNO₂ (K a = 4.5 × 10⁻⁴) with 20.0 mL of a 0.46 M NaNO₂ solution?

a. 3.71
b. 2.10
c. 2.02
d. 3.31

User Markhunte
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1 Answer

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Final answer:

To calculate the pH of a buffer solution formed by mixing HNO₂ and NaNO₂, use the Henderson-Hasselbalch equation with the concentrations and Ka value. The resulting pH is approximately 3.31.

Step-by-step explanation:

To find the pH of the solution after mixing 50.0 mL of a 0.20 M HNO₂ with 20.0 mL of a 0.46 M NaNO₂, we must consider the buffer solution formed by the weak acid (HNO₂) and its conjugate base (NO₂- from NaNO₂). Since NaNO₂ fully dissociates in water, it contributes directly to the concentration of NO₂-.

First, calculate the moles of HNO₂ and NO₂-:
HNO₂: 50.0 mL × 0.20 M = 0.0100 mol
NO₂-: 20.0 mL × 0.46 M = 0.0092 mol

Next, account for the dilution when the solutions are mixed: total volume = 50.0 mL + 20.0 mL = 70.0 mL
Moles remain the same, so the new concentrations are:
[HNO₂] = 0.0100 mol / 0.0700 L = 0.1429 M
[NO₂-] = 0.0092 mol / 0.0700 L = 0.1314 M

Since we have the concentrations and the Ka of HNO₂ (4.5 × 10-4), we can use the Henderson-Hasselbalch equation to find the pH.
pH = pKa + log([conjugate base]/[acid])
pKa of HNO₂ = -log(4.5 × 10-4) = 3.35
pH = 3.35 + log(0.1314 / 0.1429)

Calculate the log value and add to pKa to obtain the pH of the buffer solution:

pH = 3.35 + log(0.1314 / 0.1429) = 3.35 + (-0.036) = 3.314, which rounds to 3.31.

User Dilermando Lima
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