How much HOCl (Ka = 2.5x10-8) (HOCl <=> H+ + OCl- must be added to pure...

Question

Question

How much HOCl (K_a = 2.5x10^-8) (HOCl <=> H⁺ + OCl^- must be added to pure water to make a solution of pH 4.3?

Answer 1

Answer #1

Let the concentration of HOCl be c

use:

pH = -log [H+]

4.3 = -log [H+]

[H+] = 5.012*10^-5 M

HOCl dissociates as:

HOCl -----> H+ + OCl-

c 0 0

c-x x x

Ka = [H+][OCl-]/[HOCl]

Ka = x*x/(c-x)

2.5*10^-8 = 5.012*10^-5*5.012*10^-5/(c-5.012*10^-5)

c-5.012*10^-5 = 0.1005

c=0.100 M

Answer: 0.100 M

Answer 2

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