otic acid, hasa solution of a 0.120 M solution of this acid is titrated with 0.2
ID: 1036708 • Letter: O
Question
otic acid, hasa solution of a 0.120 M solution of this acid is titrated with 0.250 M NaOH, and the titration reaction is Calculate the pH of the solution at (a) the start of the titration (b) at the equivalence point. Hint: At the equivalence point, CaHsN2 O, (ag) undergoes the hydrolysis 2.(12pts) What mass of sodium formate, NaHCO2 must be dissolved in 250.0 mL of 0.1000 M formie acid,HCO2H, a weak monoprotic acid, to yield a buffer solution of pH 3.65? The Ka of formic acid is 1.7x 104. Assume that the volume change when the sodium formate is added is negligibleExplanation / Answer
ANSWER
(a) pH at the start of the titration:
At te start of the titration only acid will be present as:
C4H4N2O3 <--------> C4H3N2O3-+ H+
K = [C4H3N2O3-] [H+ ] / [C4H4N2O3 ] = x2 / (0.12 - x)
x = amount of C4H4N2O3 dissociated
9.9 X 10-5 = x2 / (0.12 - x)
9.9 X 10-5 (0.12 - x) = x2
x2 + 9.9 X 10-5 x - 1.88 X 10-5 = 0
We got a quadraitic equation the solution of which gives the value of x
x = 0.0043M
pH = -log[H+] = -log[0.0043] = 2.36
(b) At equivalence point the conjugate base of the acid will undergo hydrolysis as:
C4H3N2O3- + H2O <-------->C4H4N2O3 + OH-
We know Ka X Kb = 10-14
Kb = 10-14/ Ka
Hence Ka = 10-14/Kb = [C4H4N2O3 ] [OH-] / [C4H3N2O3-]
10-14 / 9.9 X 10-5 = x2 / (0.0043-x)
1.0 X 10-10 (0.0043-x) = x2
4.3 X 10-13 - 1.0 X 10-10x = x2
x2 + 1.0 X 10-10x - 4.3 X 10-13 = 0
x = 6.55 X 10-7
Hence [OH-] = 6.55 X 10-7
pOH = -log[OH-] = -log(6.55 X 10-7)
pOH = 6.18
pH + pOH =14
pH = 14 - pOH = 14 - 6.18 = 7.82
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