Topic: “Ka Weak Acid Calculati
Topic: “Ka Weak Acid Calculations”
1)The pH of an aqueous solution of 0.420 Mhydrofluoric acid is_______ .
2)The pOH of an aqueous solution of 0.581 Mhydrocyanic acid is_______ .
Answer:
1)
HF ————————> H+ + F-
0.420 0 0 ——–>I
-x +x +x ——– >C
0.420-x x x————–> E
Ka = [H+][F-]/[HF]
6.3 x 10^-4 = x^2 / 0.420 -x
x^2 + 6.3 x 10^-4 x – 2.65 x 10^-4 = 0
x = 0.0160 M
[H+] = 0.0160 M
pH = -log [H+]
pH = -log (0.0160)
pH = 1.80
2)
pH = 1/2 [pKa – log C]
pH = 1/ 2 [9.21 -log 0.581]
pH = 4.72
pH + pOH = 14
pOH = 9.28
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