Itzel R.

asked • 10/26/20

What is the pH of a solution of 0.300 M HNO₂ containing 0.170 M NaNO₂? (Ka of HNO₂ is 4.5 × 10⁻⁴)

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J.R. S. answered • 10/26/20

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A buffer is made because you have a weak acid (HNO2) and the conjugate base (NO2-)

Use the Henderson Hasselbalch equation:

pH = pKa + log [NO2-] / [HNO2]

pH = ?

pKa = - log Ka = -log 4.5x10-4 = 3.35

log [NO2-] / [HNO2] = log 0.170 / 0.130) = 0.12

pH = 3.35 + 0.12

pH = 3.47



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Connor C.

This seems wrong. Using Henderson-Hasselbalch: pH = pKa = -log(Ka) + log(Molarity acid/base) You get an answer of 3.10. -log(4.5*10^-4) = 3.35 3.35 + log(.170/.3) = 3.10
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04/04/24

J.R. S.

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I agree that my previous answer was incorrect because I missed the presence of the conjugate base, but I get a different answer than what you find. HH equation is pH = pKa + log (base/acid).
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04/04/24

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