Calculate the pH at each of the following points in the titration of 50.00 mL of

Question

Calculate the pH at each of the following points in the titration of 50.00 mL of a 0.2000 M acetic acid solution using 0.315 M NaOH. (Ka = 1.8x10-5)

a: Initial pH: 0 mL of base added. Enter your answer as a number with at least 2 sig figs.

b: What is the pH when you are halfway to equivalence point (you have added half as many moles of base as the starting moles of acid)? Enter your answer as a number with at least 2 sig figs.  

c: What is the pH after 23.85 mL of NaOH have been added? Enter your answer with at least 2 sig figs.

d: What is the pH at the equivalence point? Enter your answer as a number with at least 2 sig figs.  

e: What is the pH after 48.56 mL of NaOH have been added? Enter your answer with at least 3 sig figs.  

Explanation / Answer

pKa of acetic acid = 4.74

a)

pH = 1/2 (pKa - log C)

     = 1/2 (4.74 - log 0.200)

pH = 2.72

b)

At half - equivalence point :

pH = pKa

pH = 4.74

c)

millimoles of acetic acid = 50 x 0.2 = 10

millimoles of NaOH = 23.85 x 0.315 = 7.513

CH3COOH +   NaOH   -----------> CH3COONa   + H2O

    10               7.513                          0                    0

2.487                 0                          7.513

pH = pKa + log [salt / acid]

     = 4.74 + log [7.513 / 2.487]

pH = 5.22

d)

volume of NaOH = 31.75

salt concentration = 10 / 50 + 31.75 = 0.122 M

pH = 7 +1/2 (pKa + log C)

     = 7 + 1/2 (4.74 + log 0.122)

pH = 8.91

d)

[OH-] = 0.0537 M

pOH = 1.27

pH = 12.73

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