Consider the titration of 30.00 mL of an HCl solution with an
unknown molarity. The volume of 0.1200 M NaOH required to reach the
equivalence point was 42.50 mL.
a. Write the balanced chemical equation for the neutralization reaction.
b. Calculate the molarity of the acid.
c. What is the pH of the HCl solution before any NaOH is
d. What is the pH of the analysis solution after exactly 42.25 mL of NaOH is added? e. What is the pH at the equivalence point?
f. What is the pH of the analysis solution after exactly 42.55 mL of NaOH is added?
NaOH + HCl ----------------> NaCl + H2O
at equivalence point :
millimoles of HCl = millimoles of NaOH
30 x M = 42.50 x 0.1200
M = 0.1700 M
molarity of acid = 0.1700 M
pH = -log [H+] = -log (0.17)
pH of HCl = 0.77
moles of NaOH = 42.25 x 0.12 = 5.07
moles of HCl = 30 x 0.17 = 5.1
concentration [H+]= moles of HCl - moles of NaOH / total volume
= 5.1 - 5.07 / (30 + 42.25)
= 4.15 x 10^-4 M
pH = -log [H+] = -log (4.15 x 10^-4)
pH = 3.38
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