Question

A weak monoprotic acid, HA dissociates by 0.402 % and has a pH of 2.36. Calculate the Ka value for the acid HA. Record your answer in scientific notation to 3 sig figs.

At 272.8 ^oC, the K_eq for the reaction:

1 A + 3 B ↔↔ 4 AB

is 3.41⋅1023.41⋅102

What is K_p? Enter your answer in scientific notation.

Answer 1

1)

use:

pH = -log [H+]

2.36 = -log [H+]

[H+] = 4.365*10^-3 M

HA   <—>   H+   +   A-

c       0       0   (initial)

c-x       x       x   (at equilibrium)

x = [H+] = 4.365*10^-3 M

% dissociation = x*100/c

0.402 = (4.365*10^-3)*100/c

c = 1.086 M

Ka = [H+][A-]/[HA]

= x*x / (c-x)

= (4.365*10^-3)*(4.365*10^-3)/(1.086 - 4.365*10^-3)

= 1.76*10^-5

Answer: 1.76*10^-5

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