a)
CH3COOH dissociates as:
CH3COOH -----> H+ + CH3COO-
0.1 0 0
0.1-x x x
Ka = [H+][CH3COO-]/[CH3COOH]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.8*10^-5)*0.1) = 1.342*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
1.8*10^-5 = x^2/(0.1-x)
1.8*10^-6 - 1.8*10^-5 *x = x^2
x^2 + 1.8*10^-5 *x-1.8*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.8*10^-5
c = -1.8*10^-6
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 7.2*10^-6
roots are :
x = 1.333*10^-3 and x = -1.351*10^-3
since x can't be negative, the possible value of x is
x = 1.333*10^-3
use:
pH = -log [H+]
= -log (1.333*10^-3)
= 2.8753
Answer: 2.88
b)
Given:
M(CH3COOH) = 0.1 M
V(CH3COOH) = 25 mL
M(NaOH) = 0.12 M
V(NaOH) = 15 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.1 M * 25 mL = 2.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.12 M * 15 mL = 1.8 mmol
We have:
mol(CH3COOH) = 2.5 mmol
mol(NaOH) = 1.8 mmol
1.8 mmol of both will react
excess CH3COOH remaining = 0.7 mmol
Volume of Solution = 25 + 15 = 40 mL
[CH3COOH] = 0.7 mmol/40 mL = 0.0175M
[CH3COO-] = 1.8/40 = 0.045M
They form acidic buffer
acid is CH3COOH
conjugate base is CH3COO-
Ka = 1.8*10^-5
pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.745+ log {4.5*10^-2/1.75*10^-2}
= 5.155
Answer: 5.16
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