Ka of HN3 = 2.5×10–5
concentration of HN3 = 0.159 M
HN3 --------------> N3- + H+
0.159 0 0
0.159 - x x x
Ka = x^2 / 0.159 - x
2.5 x 10^-5 = x^2 / 0.159 - x
x = 1.98 x 10^-3
[HN3] = 0.157 M
[H3O+] = 1.98 x 10^-3 M
[N3-] = 1.98 x 10^-3 M
pH = 2.70
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Mo B 000 A 2L 3 /3
-1 + i 3 3 ,find the following (a) (b) y'Y
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