consider the titration of 23 mL sample of 0.100 M HC2H3O2 with .130 M NaOH. Dete

Question

consider the titration of 23 mL sample of 0.100 M HC2H3O2 with .130 M NaOH. Determine the initial pH, the volume of added base required to reach the equivalence point. The pH at 5.00 mL of added base. The pH at one half of the equivalence point and the pH at the equivalence point onsider the titration of a 23.0-mL sample of 100 M HC2H3O2 with 0.130 M NaOH etermine each of the following. the initial pH Express your answer using two decimal places. pH Submit My Answers Give Up Part B the volume of added base required to reach the equivalenc Submit My Answers Give Up Part C the pH at 5.00 mL of added base Express your answer using two decimal places. Suba nit My Answers Give Up

Explanation / Answer

a)

initial pH:

Ka = [H+][A-]/[HA]

1.8*10^-5 = x*x/(M-x)

1.8*10^-5 = x*x/(0.1-x)

x = 0.001332

pH= -log(0.001332)

pH = 2.875

B)

Volume for equivalence will be

mmol of acid = mmol of base

M1*V1 = M2*V2

0.1*23 = 0.13*V2

V2 = 0.1*23/0.13 = 17.6923 mL of base required

C)

find pH when

V = 5 mL of base is added

so

mmol of acid = MV = 23*0.1 = 2.3 mmol of acid

mmol of base = MV = 0.13*5 = 0.65 mmol

acid reacts to form conjguate so

mmol of acid left = 2.3-0.65 = 1.65

mmol of conjugate = 0 + 0.65 = 0.65

then

this is abuffer so

pH = pKa + log(A-/HA)

pH = 4.75 + log(0.65/1.65)

pH = 4.34

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