Calculate the pH of 0.010 M Ca(CH3CO2)2 (calcium acetate) solution.
A. 2.00
B. 5.47
C. 5.62
D. 8.37
E. 8.52
[CH3CO2-] = 2*[Ca(CH3CO2)2]
= 2*0.010 M
= 0.020 M
Ka of CH3CO2H is 1.8*10^-5
lets find Kb for CH3CO2-
use:
Kb = Kw/Ka
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/1.8*10^-5
Kb = 5.556*10^-10
CH3CO2- dissociates as
CH3CO2- + H2O -----> CH3CO2H + OH-
0.02 0 0
0.02-x x x
Kb = [CH3CO2H][OH-]/[CH3CO2-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((5.556*10^-10)*2*10^-2) = 3.333*10^-6
since c is much greater than x, our assumption is correct
so, x = 3.333*10^-6 M
use:
pOH = -log [OH-]
= -log (3.333*10^-6)
= 5.48
use:
PH = 14 - pOH
= 14 - 5.48
= 8.52
Answer: 8.52
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Please explain to me how I solve this
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