Titration Curves: Strong Acid+ Strong Base 15. Consider the titration of the str

Question

Titration Curves: Strong Acid+ Strong Base 15. Consider the titration of the strong acid HCl with the strong base NaOH. Calculate the pH values at the following points in the titration. Be sure to supplement your calculations with the chemical equations describing the chemistry of each region of the titration curve. The initial pH of a 40.00 mL sample of 0.1000M HCI The pH of the sample after the addition of 20.00 mL (total) of 0.1000 M NaOH The pH of the sample at the equivalence point The pH of the sample after the addition of 50.00 mL(total) of 0.1000M NaOH a. b. c. d.

Explanation / Answer

(a) pH = - log [ H+] = -log (0.1000) = 1

(b) 40 ml of 0.1000 M HCl contains 40×0.1/1000 mole HCl = 0.004 mole HCl

20 ml of 0.1000 M NaOH contains 20×0.1000/1000 mole NaOH = 0.002 mole NaOH

0.002 mole NaOH neutralises 0.002 mole HCl and rest (0.004-0.002) mole = 0.002 mole HCl remains unreacted.

Total volume of mixture = (40+20) ml. = 60 ml.= 0.06 liters

Concentration of excess HCl in solution : 0.002 mole/ 0.06 lit = 0.033 M

pH =-log [H+] = -log (0.033)= 1.48

c) This is a titration of syrong acid versus strong base. At equivalence point pH =7.0

d)

40 ml of 0.1000 M HCl solution contains 0.004 mole HCl

50 ml of 0.1000 M NaOH contains 50 ×0.1000/1000 mole NaOH = 0.005 mole NaOH

So; excess moles of NaOH in solution = 0.006-0.004 mole =0.001 mole

Total volume = 40+50 ml =90 ml = 0.09 lit

Conc. of excess NaOH = 0.001/0.09 M = 0.011 M

pOH = -log [OH-] = -log (0.011)= 1.96

pH =14-pOH = 14-1.96 = 12.04

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