Answer
pH = 8.64
concentration of propionic acid = 4.38×10-6M
Explanation
C3H5O2- ion is partly hydrolysed by water
C3H5O2- (aq) + H2O(l) <--------> HC3H5O2(aq) + OH-(aq)
Kb = [HC3H5O2][OH-]/[C3H5I2-]
Kb = Kw /Ka
Kw = ionic product of water, 1.00×10-14
Kb = 1.00×10-14/1.3 ×10-5 = 7.69×10-10
Initial concentration
[C3H5O2-] = 0.025
[HC3H5O2] = 0
[OH-] = 0
change in conctration
[C3H5O2-] = -x
[HC3H5O2] = +x
[OH-] = +x
Equillibrium concentration
[C3H5O2-]= 0.025 - x
[HC3H5O2] = x
[OH-] = x
so,
x2/(0.025 - x) = 7.69×10-10
we can assume , 0.025 - x = 0.025 because x is small value
x2 /0.025 = 7.69×10-10
x2 = 1.92 ×10-11
x = 4.38 × 10-6
Therefore
[OH-] = 4.38×10-6M
[HC3H5O2] = 4.38×10-6M
pOH = -log[OH-] = -log(4.38 ×10-6) = 5.36
pH = 14 - pOH
pH = 14 - 5.36
pH = 8.64
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