Question

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

(α_OBr- ≈ 350 pm)

Answer 1

Ka = Activity of H+ * Activity of OBr- /(Activity of HOBr)

Ka = 2*10^-9

Activity = Y-x * [X]

Ka = Activity of H+ * Activity of OBr- /(Activity of HOBr)

2*10^-9 = Y-H * [H+] * Y-OBr- * [OBr] / [HOBr]

IS = 0.05

i)

Recall that:

-log(γ) = 0.51*(Zi^2)*sqrt(I.S.) / ( 1 + (α * sqrt(I.S)/305))

Where

γi = activity coefficient for species “i”

αi = theoretical diameter in pm (10^-12 m)

Zi = Charge of ion

I.S. = ionic Strength (usually used as μ as well)

If we wanted only γ

γ = 10^-(0.51*(Zi^2)*sqrt(I.S.) / ( 1 + (α * sqrt(I.S)/305)))

for H+, alpha = 900, +1

γ = 10^-(0.51*(1^2)*sqrt(0.05) / ( 1 + (900 * sqrt(0.05)/305))) = 0.85367

for OBr-

γ = 10^-(0.51*(1^2)*sqrt(0.05) / ( 1 + (350 * sqrt(0.05)/305))) = 0.8114

2*10^-9 = Y-H * [H+] * Y-OBr- * [OBr] / [HOBr]

2*10^-9 = 0.85367 * [H+] * 0.8114 * [OBr] / [HOBr]

(2*10^-9)/(0.85367 *0.8114 ) =  [H+] * [OBr] / [HOBr]

2.887*10^-9 = x*x/(0.001-x)

x = sqrt((2.887*10^-9)*0.001)

x = 0.00000169

[H+] = 0.00000169

Activity of H+ = 0.00000169*0.85367 = 0.00000144

pH = -log(Activit H+) = -log(0.00000144)= 5.8416