Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to...

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Assume you dissolve 0.235 g of the weak acid benzoic acid,C6H5CO2H, in enough water to make 3.00 âœ• 102 mL of solution andthen titrate the solution with 0.128 M NaOH.
C6H5CO2H(aq) + OH-(aq) C6H5CO2-(aq) + H2O(â„“)
What are the concentrations of the following ions at theequivalence point?
Na+, H3O+, OH-
C6H5CO2-
_M Na+
_M H3O+
_M OH-
_M C6H5CO2-
What is the pH of the solution?

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4.0 Ratings (614 Votes)

no of moles of benzoic acid = W/G.M.Wt

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.235/122 = 0.001924 moles

Â Â no of moles of NaOHÂ Â = molarity * volume in L

Â Â Â Â Â Â Â Â Â 0.001924Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.128* volume in L
volume in LÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.001924/0.128 = 0.015L

total volumeÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.015 + 0.3 = 0.315L
molarity of C6H5COO-Â Â = no of moles/volume

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.001924/0.315Â Â = 0.0061 M
Â Â Â Â Â Â C6H5COO^- + H2O ---------> C6H5COOH + OH^-
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â kbÂ Â = x*x/0.0061-x

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1.63*10^-10Â Â = x²/0.0061-x

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1.63*10^-10*(0.0061-x)Â Â = x²

^{Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â} x = 9.97*10^-7

Â Â Â Â Â Â Â Â Â Â Â Â Â Â [OH-] = x = 9.97*10^-7 M
Â Â Â Â Â Â Â Â Â Â Â Â Â Â [C6H5COO-] = 0.0061-0.000000997 = 0.00609M
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [H3O+] = Kw/[OH-]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1*10^-14/9.97*10^{-7Â Â =} 1*10^-7
M

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Na+]Â Â = 0.0061M

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Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to...

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