Answer:
Concentration of hydroxide-ion at equivalence point = [tex]8.3\times 10^{-6}M[/tex]
Explanation:
[tex]HC_{3}H_{5}O_{2}+KOH\rightarrow C_{3}H_{5}O_{2}^{-}K^{+}+H_{2}O[/tex]
1 mol of [tex]HC_{3}H_{5}O_{2}[/tex] reacts with 1 mol of KOH to produce 1 mol of [tex]C_{3}H_{5}O_{2}^{-}[/tex]
At equivalence point, all [tex]HC_{3}H_{5}O_{2}[/tex] gets converted to [tex]C_{3}H_{5}O_{2}^{-}[/tex].
Moles of [tex]C_{3}H_{5}O_{2}^{-}[/tex] produced at equivalence point is equal to moles of KOH added to reach equivalence point.
So, moles of [tex]C_{3}H_{5}O_{2}^{-}[/tex] produced = [tex]\frac{43.76\times 0.141}{1000}moles=0.00617moles[/tex]
Total volume of solution at equivalence point = (25.00+43.76) mL = 68.76 mL
Concentration of [tex]C_{3}H_{5}O_{2}^{-}[/tex] at equivalence point = [tex]\frac{0.00617\times 1000}{68.76}M=0.0897M[/tex]
[tex]OH^{-}[/tex] produced at equivalence point is due to hydrolysis of [tex]C_{3}H_{5}O_{2}^{-}[/tex]. We have to construct an ICE table to calculate concentration of [tex]OH^{-}[/tex] at equivalence point.
[tex]C_{3}H_{5}O_{}^{-}+H_{2}O\rightleftharpoons HC_{3}H_{5}O_{2}+OH^{-}[/tex]
I:0.0897 0 0
C: -x +x +x
E: 0.0897-x x x
[tex]\frac{[HC_{3}H_{5}O_{2}][OH^{-}]}{[C_{3}H_{5}O_{2}^{-}]}=K_{b}(C_{3}H_{5}O_{2}^{-})=\frac{10^{-14}}{K_{a}(HC_{3}H_{5}O_{2})}[/tex]
species inside third bracket represent equilibrium concentrations
So, [tex]\frac{x^{2}}{0.0897-x}=7.69\times 10^{-10}[/tex]
or, [tex]x^{2}+(7.69\times 10^{-10}\times x)-(6.90\times 10^{-11})=0[/tex]
So, [tex]x=\frac{-(7.69\times 10^{-10})+\sqrt{(7.69\times 10^{-10})^{2}+(4\times 6.90\times 10^{-11})}}{2}[/tex]M = [tex]8.3\times 10^{-6}M[/tex]
So, concentration of hydroxide-ion at equivalence point = x M = [tex]8.3\times 10^{-6}M[/tex]
