pH=8.87
Reaction
C₂H₄O₂+NaOH⇒CH₃COONa+H₂O
at the equivalence point = mol C₂H₄O₂= mol NaOH
mol C₂H₄O₂ : 50 x 0.1 = 0.5 mlmol=5.10⁻⁴ mol
The two reactants have completely reacted, and there is only salt(CH₃COONa) and water(H₂O), there will be hydrolysis
For acids from weak acids and strong bases (the solution is alkaline) then the calculation:
[tex]\tt [OH^-]=\sqrt{\dfrac{Kw}{Ka}\times M }[/tex]
M=anion concentration=CH₃COO⁻
Ka=acid constant(for CH₃COOH,Ka=1.8.10⁻⁵)
[tex]\tt [OH^-]=\sqrt{\dfrac{10^{-14}}{1.8.10^{-5}}\times 0.1 }[/tex]
[tex]\tt [OH^-]=\sqrt{5.6.10^{-11}}=7.483\times 10^{-6}[/tex]
pOH=6-log 7.483=5.13
pH= 14 - 5.13=8.87
The pH of the solution is the negative log [H] concentration. The pH of the given solution is 8.87 at the equivalence point.
The given Reaction is:
C₂H₄O₂+ NaOH ⇒ CH₃COON + H₂O
The pH of the solution can be calculated by,
[tex]\bold{pH = 14 - pOH}[/tex]................1
So first, calculate for pOH
[tex]\bold {pOH = \sqrt {\dfrac {Kw}{Ka} \times M}}[/tex]
Where,
Ka - dissociatioon constant = 50 x 0.1 = 0.5 ml = 5x10⁻⁴ mol
Kw - concentaion water ionic products = [tex]\bold {10^{-14}}[/tex]
M - anion concentration( CH₃COO⁻) = 0.1 M
Put the values in the formula,
[tex]\bold {pOH = \sqrt {\dfrac {\bold {10^{-14}}}{5\times 10^{-4} mol} \times 0.1}}\\\\\bold {pOH = 7.483 \times 10^{-6}}[/tex]
Since the pOH is the negative log of [OH⁻] concentration,
pOH = 6 -ln 7.483
pOH = 5.13
Put the value of pOH in the first equation,
pH= 14 - 5.13
pH = 8.87
Therefore, the pH of the given solution is 8.87 at the equivalence point.
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