Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 5.00 ✕ 102 mL of solution and then titrate the solution with 0.128 M NaOH.
C6H5CO2H(aq) + OH-(aq) C6H5CO2-(aq) + H2O(ℓ)
What are the concentrations of the following ions at the equivalence point?
Na+, H3O+, OH-
C6H5CO2-
_____ M Na+
_____ M H3O+
_____ M OH-
_____ M C6H5CO2-
What is the pH of the solution?
pH =
please help if you can :)
1)
moles of benzoic acid = 0.235 / 122.12 = 1.924 x 10^-3 mol
concentration of benzoic acid = 1.924 x 10^-3 / 0.500
= 3.85 x 10^-3 M
At equivalence point :
moles of acid = moles of NaOH
1.924 x 10^-3 = 0.128 x V
V = 0.015 L
total volume = 0.015 + 0.500 = 0.515 L
here salt only remains.
[C6H5COO-] = 1.924 x 10^-3 / 0.515 = 0.00374 M
C6H5COO- + H2O ------------> C6H5COOH + OH-
0.00374 0 0
0.00374 - x x x
Kb = x^2 / 0.00374 - x
1.59 x 10^-10 = x^2 / 0.00374 - x
x = 7.705 x 10^- 7
[OH-] = 7.705 x 10^-7 M
pOH = 6.11
pH = 7.89
0.128 M = Na+
1.30 x 10^-8 M = H3O+
7.71 x 10^-7 M = OH-
0.00374 M = C6H5CO2-
pH = 7.89
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