Question

“Calculate the pH of 0.5 mM ammonia in aqueous solution at 25 °C. The pKa of the ammonium cation is 9.25. Allowed error is 0.03 pH units. State why the result offered by WolframAlpha was wrong.

Answer 1

pka of NH4+ is 9.25

we have relation pka of acid ( NH4+) + pkb of its conjugate base (NH3) = 14

pkb of NH3 ( conjugate base of NH4+) = 14-pka of NH4+ = 14- 9.25 = 4.75

Kb = 10^ - pkb = 10^ -4.75 = 1.78 x 10^ -5

Now base dissociation eq is given by

NH3 (aq) + H2O (l) <---> NH4+ (aq) + OH- (aq)

Initially [NH3]] = 0.5 mM = 0.0005 M , [NH4+] =[OH-] = 0

at equilibrium [NH3] = 0.0005-X , [NH4+] =[OH-] = X

Kb = [ NH4+] [OH-] / [NH3]

1.78 x 10^ -5 = ( X^2) / ( 0.0005-X)

X^2 + ( 1.78 x 10^ -5) X - (8.9 x 10^ -9 ) = 0

solving this quadratic eq we get X = [OH-] = 0.000086

pOH = -log [OH-] = -log ( 0.000086) = 4.0655

pH = 14 - 4.0655 = 9.93