The Ka of propanoic acid (C2H5COOH) is 1 34 10 5 Calculatethe pH of ...

C2H5COOH C2H5COO H 0551 0 0 initial 0551x x x at equilibrium K ....

The Ka of propanoic acid (C2H5COOH) is 1.34 Ã— 10-5. Calculate the pH of the solution...

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The Ka of propanoic acid (C2H5COOH) is 1.34 Ã— 10-5. Calculatethe pH of the solution and the concentrations of C2H5COOH andC2H5COOâ€“ in a 0.551 M propanoic acid solution at equilibrium.

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4.4 Ratings (638 Votes)

C2H5COOH ---> C2H5COO-Â Â +Â Â H+
0.551Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â (initial)
0.551-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â (at equilibrium)

Ka = [C2H5COO-][H+] / [C2H5COOH ]
1.34*10^-5 = x*x / (0.551-x)

Here Ka is very small, so x will be very small and it can be ignored as compared to 0.551
So, above expression becomes
1.34*10^-5 = x*x / 0.551
x=2.72*10^-3 M

Answer:
[C2H5COOâ€“] = x = 2.72*10^-3 M
[C2H5COOH ] = 0.551-x= 0.551-2.72*10^-3 = 0.548 M

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