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Mashcka [7]
3 years ago
11

A nitrous acid buffer solution contains 0.85 M nitrous acid (HNO2) and 0.61 M of its conjugate base (NO2−). If a chemist adds 0.

15 mol of hydrobromic acid (HBr), a strong acid, to 0.50 L of the buffer solution, what will be the final pH of the solution? (The pKa of HNO2 is 3.25. Assume the volume of the added HBr is negligible.)
Chemistry
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

2.68

Explanation:

At the solution, the number of moles of each substance (acid and conjugate base) is the volume multiplied the concentration

nHNO₂ = 0.50 L * 0.85 mol/L = 0.425 mol

nNO₂⁻ = 0.50 L * 0.61 mol/L = 0.305 mol

At the buffer, the substances are in equilibrium. When HBr is added, it dissociantes in H⁺ and Br⁻, and the H⁺ will react with NO₂⁻ to form more HNO₂. So, NO₂⁻ will be consumed and HNO₂ will be formed at a 1:1:1 reaction:

nH⁺ = nHBr = 0.15 mol

nNO₂⁻ = 0.305 - 0.15 = 0.155 mol

nHNO₂ = 0.425 + 0.15 = 0.575 mol

The pH of a buffer can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[A⁻]/[HA]

Where [A⁻] is the concentration of the conjugate base, and [HA], the concentration of the acid. Because the volume is the same, it can be used the number of moles:

pH = 3.25 + log (0.155/0.575)

pH = 2.68

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