Determine the pH of the following solution:
0.852 M LiC2H3O2
C2H3O2- (acetate ion) partly hydrolysed by water
C2H3O2-(aq) + H2O(l) <--------> HC2H3O2(aq) + OH-(aq)
Kb = [HC2H3O2][OH-] / [C2H3O2-] = 5.56 ×10-10
Initial concentration
[C2H3O2- ] = 0.852
[HC2H3O2] = x
[ OH-] = x
change in concentration
[C2H3O2-] = -x
[HC2H3O2] = +x
[OH-] = + x
equilibrium concentration
[C2H3O2-] = 0.852 - x
[HC2H3O2] = x
[OH-] = x
so,
x2/( 0.852 - x) = 5.56 ×10-10
we can assume 0.852 - x = 0.852 because x is small value
x2/ 0.852 = 5.56 ×10-10
x2 = 4.737 × 10-10
x = 2.176 × 10-5
[OH-] = 2.176 ×10-5M
pOH = -log[OH-]
pOH = - log( 2.176 ×10-5)
pOH = 4.66
pH = 14 - 4.66
pH = 9.34
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