In another experiment, the student titrated 50.0mL of 0.100MHC2H3O2 with 0.100MNaOH(aq) . Calculate the pH of the solution at the equivalence point.

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,

................1

So first, calculate for pOH

Where,

Ka - dissociatioon constant = 50 x 0.1 = 0.5 ml = 5x10⁻⁴ mol

Kw - concentaion water ionic products =

M - anion concentration( CH₃COO⁻) = 0.1 M

Put the values in the formula,

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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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:

M=anion concentration=CH₃COO⁻

Ka=acid constant(for CH₃COOH,Ka=1.8.10⁻⁵)

pOH=6-log 7.483=5.13

pH= 14 - 5.13=8.87

c is the answer because when something is left after something evaporates that shows the chemical change.

b

explanation:

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