A titration is performed by adding 0.317 M KOH to 40 mL of 0.228 M HNO3. a) Calc

ID: 957319 • Letter: A

Question

A titration is performed by adding 0.317 M KOH to 40 mL of 0.228 M HNO3.

a) Calculate the pH before addition of any KOH.

b) Calculate the pH after the addition of 5.75, 14.39 and 27.77 mL of the base.(Show your work in detail for one of the volumes.)

c) Calculate the volume of base needed to reach the equivalence point.

d) Calculate the pH at the equivalence point.

e) Calculate the pH after adding 5.00 mL of KOH past the equivalence point.

f) Plot pH (y axis) versus volume of KOH added (x axis) for each calculation above. Sketch the titration curve.

a.) HNO3 is a strong acid and it will complete dissociate according to the equation

HNO3(aq) H+(aq) + NO3–(aq)

Initial: 0.350 M 0 0

[H+] = 0.228 M

pH = 0.642

b.) after Adding 5.750 mL of 0.250 M KOH

HNO3(aq) + KOH(aq) H2O(l) + KNO3(aq)

initial 0.00912 mol 0.0018 mol – 0

final 0.00732mol 0 – 0.0018 mol

[H+]= .00732/.04575 = .16 mol/L

pH = .795

similarly

for 14.39 mL of base added

[H+]= .046/.039 = .1179 mol/L

pH = .928

for 27.77 mL of base added

[H+]= .0003/.0528 = .00576 mol/L

pH = 2.2395

c.) MacidVacid = MbaseVbase

0.228 M . 40.0 mL = 0.317 M . Vbase

Vbase = 28.769 mL

d.) Add 28.769 mL of 0.250 M KOH

HNO3(aq) + KOH(aq) H2O(l) + KNO3(aq)

initial 0.00912 mol 0.00912 mol – 0

final 0 0 – 0.00912 mol

KNO3 is the salt of a strong acid and strong base, so the pH = 7.00 at the equivalence point.

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