Question

What is the pH of a 0.00100 F solution of hypobromous acid (HOBr) in 0.0500F NaCl?

Answer 1

HOBr <--> H+ + OBr-

First, assume the acid:

HF

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

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

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.001 M; then

x^2 + (2*10-9)x - ^0.001*(2*10^-9) = 0

solve for x

x =4.46*10^-7

substitute

[H+] = 0 + 4.46*10^-7= 0.01434 M

pH = -log(H+) = -log(4.46*10^-7) = 6.35