use:
pKa = -log Ka
3.6 = -log Ka
Ka = 2.512*10^-4
find the volume of NaOH used to reach equivalence point
M(H2CO3)*V(H2CO3) =M(NaOH)*V(NaOH)
0.4557 M *70.0 mL = 0.8352M *V(NaOH)
V(NaOH) = 38.1932 mL
Given:
M(H2CO3) = 0.4557 M
V(H2CO3) = 70 mL
M(NaOH) = 0.8352 M
V(NaOH) = 38.1932 mL
mol(H2CO3) = M(H2CO3) * V(H2CO3)
mol(H2CO3) = 0.4557 M * 70 mL = 31.899 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.8352 M * 38.1932 mL = 31.899 mmol
We have:
mol(H2CO3) = 31.899 mmol
mol(NaOH) = 31.899 mmol
31.899 mmol of both will react to form HCO3- and H2O
HCO3- here is strong base
HCO3- formed = 31.899 mmol
Volume of Solution = 70 + 38.1932 = 108.1932 mL
Kb of HCO3- = Kw/Ka = 1*10^-14/2.512*10^-4 = 3.981*10^-11
concentration ofHCO3-,c = 31.899 mmol/108.1932 mL = 0.2948M
HCO3- dissociates as
HCO3- + H2O -----> H2CO3 + OH-
0.2948 0 0
0.2948-x x x
Kb = [H2CO3][OH-]/[HCO3-]
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 ((3.981*10^-11)*0.2948) = 3.426*10^-6
since c is much greater than x, our assumption is correct
so, x = 3.426*10^-6 M
[OH-] = x = 3.426*10^-6 M
use:
pOH = -log [OH-]
= -log (3.426*10^-6)
= 5.4652
use:
PH = 14 - pOH
= 14 - 5.4652
= 8.5348
Answer: 8.53
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