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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
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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