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
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