no of moles of CH3COOH = molarity * volume in L
= 0.1*0.05 = 0.005moles
no of moles of KOH = molarity * volume in L
= 0.1*0.05 = 0.005moles
total volume = 50 +50 = 100ml = 0.1L
CH3COOH(aq) + KOH(aq) ----------> CH3COOK(aq) + H2O(l)
I 0.005 0.005 0
C -0.005 -0.005 0.005
E 0 0 0.005
molarity of CHCOOK = no of moles/volume in L
= 0.005/0.1 = 0.05M
CH3COO^- (aq) + H2O(l) ----------> CH3COOH(aq) + OH^- (aq)
I 0.05 0 0
C -x +x +x
E 0.05-x +x +x
Kb = Kw/ka
= 1*10^-14/(1.8*10^-5) = 5.6*10^-10
Kb = [CH3COOH][OH^-]/[CH3COO^-]
5.6*10^-10 = x*x/(0.05-x)
5.6*10^-10*(0.05-x) = x^2
x = 5.3*10^-6
[OH^-] = x = 5.3*10^-6M
POH = -log[OH^-]
= -log(5.3*10^-6)
= 5.2757
PH = 14-POH
= 14-5.2757
= 8.72
b. 8.72 >>>answer
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