Question

Calculate the pH of 250 mL of a 0.46 M solution of HN3 before and after the addition of 0.076 mol of N3 Ka(HN3) = 0.000019 initial pH = final pH =

Answer 1

1)

HN3 dissociates as:

HN3 -----> H+ + N3-

0.46 0 0

0.46-x x x

Ka = [H+][N3-]/[HN3]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.9*10^-5)*0.46) = 2.956*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.956*10^-3 M

So, [H+] = x = 2.956*10^-3 M

use:

pH = -log [H+]

= -log (2.956*10^-3)

= 2.5292

Answer: 2.53

2)

[N3-] = mol of N3- / volume in L

= 0.076 mol / 0.250 L

= 0.304 M

Ka = 1.9*10^-5

pKa = - log (Ka)

= - log(1.9*10^-5)

= 4.721

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.721+ log {0.304/0.46}

= 4.541

Answer: 4.54