I\'m trying to find the Volume, PH and PKa for (1/4) equivalence point (1/2) equ

Question

I'm trying to find the Volume, PH and PKa for (1/4) equivalence point (1/2) equivalence point and at the 3/4 equivalence point. Molarity of NaOH used is 0.05131. 10.0 mL of unknown added to a flask with aprox 40 ml of water. After i titrated the weak acid with the strong base i plotted my point and found the equivalence point to be V=7.82 mL and a ph of 8. SO moles of acid is = to moles of base at equivalence point so moles was determined to be 0.0004012 moles and the molarity of unknown is 0.04012 M unknow acid. I know at 1/2 equivalence point ph=pka bc log 1 =1 for the henderson haselbach equation. But i'm stuck finding the pka and everything else. Please show all work

Explanation / Answer

acid + NaOH

molarity of acid found = 0.04012 M

pH at equivalence point = 8

pH = -log[H+] = 8

[H+] = 1 x 10^-8 M

[OH-] = 1 x 10^-14/1 x 10^-8 = 1 x 10^-6 M

at equivalence point,

Kb = [HA][OH-]/[A-]

     = (1 x 10^-6)^2/(0.04012)

     = 2.5 x 10^-11

Ka = 1 x 10^-14/2.5 x 10^-11 = 4.01 x 10^-4

pKa = -log[Ka] = 3.4

so pH at half-equivalence = 3/4

at 3/4 equivalence point

volume of base added = 3 x 7.82/4 = 5.865 ml

[A-] formed = 0.05131 M x 5.865 ml/55.865 ml = 0.0054 M

[HA] remained = (0.04012 M x 50 ml - 0.05131 M x 5.865 ml)/55.865 ml = 0.0305 M

pH = pKa + log(A-/HA)

     = 3.4 + log(0.0054/0.0305)

     = 2.65

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