The pH of a solution of HN3 (Ka = 1.9 x10^-5) and NaN3 is 4.86. What...

Question

Question

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?

Answer 1

Answer #1

pKa = -log Ka = -log(1.9 x 10^-5) = 4.72

HN3 + NaN3 mixer can act as acidic buffer

For acidic buffer

Henderson-Hasselbalch equation

pH = pKa + log[salt/acid]

pH = pKa + log[NaN3/HN3]

4.86 = 4.72 + log[NaN3/0.016]

[NaN3] = 0.022

molarity of NaN3 = 0.022M

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Answer 2

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