Question

Calculate the Ka, for an unknown acid HX, given that a 0.25M salt solution NaX has a pH of 8.48

answer: 2.7x10^-4

Answer 1

pOH = 14 - pH

= 14 -8.48 = 5.52

OH- = 10^-5.52

= 3.02*10^-6 M

NaX = Na+(aq) + X-

NaX = X- = 0.25

X-(aq) + H2O = HX + OH-

0.25 - 0 0 ( initially )

(0.25-x) x x (equilibrium)

But x = OH- = HX = 3.02*10^-6

X- = ( 0.25 - (3.02*10^-6) = 0.25 appx

Kb= ( HX) * (OH-) / ( X-)

= ( 3.02*10^-6)*(3.02*10^-6) / 0.25

= 3.65*10^-11

Ka = 10^-14 / kB

=10^-14 / (3.65*10^-11 )

= 2.7411*10^-4

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