eight Consider the titration of a 35.0ml sample of 0.175 M HBr with .205M KOH. D

Question

eight

Consider the titration of a 35.0ml sample of 0.175 M HBr with .205M KOH. Determine the following.

partA the initial pH express answer using three decimal places

PartB the volume of added base required to reach the equivalence point express answer in millimeters

PartC the pH at 11.9mL of added base express answer using three decimal places

PartD the pH at the equilalence point express answer as a whole number

PartE the pH after adding 5.0 mL of base beyond the equilance point express your answer using two decimal places

Explanation / Answer

a)

initially:

[H+] = 0.175 M

pH = -log(0.175)

pH = 0.75

b)

Vequivalence:

Macid*Vacid = Mbase*Vbase

Vbase = Macid*Vacid/Mbase = 0.175*35/0.205 = 29.878 mL

c)

mmol of acid = MV = 0.175*35 = 6.125

mmol of base = MV = 11.9*0.205 = 2.4395

after reaction

mmol of H+ = 6.125-2.4395 = 3.6855

[H+] = mmol/mL = 3.6855/(35+11.9) = 0.07858

d)

in equivalence pH = 7 since strong acid/base

e)

mmol of OH- = (5)(0.205)/(29.878 +5+35) = 0.01466

ph = !4 + log(OH) = 14 + log(0.01466) = 12.17

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