9) no of mol of formicacid = 50*0.05 = 2.5 mmol no of mol of NaOH required to re
ID: 699622 • Letter: 9
Question
9) no of mol of formicacid = 50*0.05 = 2.5 mmol
no of mol of NaOH required to reach equivalence point = 2.5 mmol
volume of NAoh = 2.5/0.05 = 50 ml
concentration of salt = 2.5/(100) = 0.025 M
pH = 7+1/2(pka+logC)
pka of formic acid = 3.745
= 7+1/2(3.745+log0.025)
= 8.07
10) pka of aceticacid =-logka
= -log(1.8*10^-5)
= 4.745
pH of acetic acid = 1/2(pka-logC)
pka = 4.745
C = concentration = 0.1 M
pH = 1/2(4.745-log0.1) = 2.87
b)
no of mol of acetic acid = 75*0.1 = 7.5 mmol
no of mol of NaOH must add = 25*0.1 = 2.5 mmol
pH = pka + log(base/acid-base)
pka = 4.745
= 4.745+log(2.5/(7.5-2.5))
= 4.44
c) at half equivalence point, pH = pka = 4.745
d) at equivalence point
no of mol of acetic acid = 75*0.1 = 7.5 mmol
no of mol of NaOH must add = 7.5 mmol
volume of NaOH must add = n/M = 7.5/0.1 = 75 ml
concentration of salt = n/V = 7.5/(75+75) = 0.05 M
pH = 7+1/2(pka+logC)
pka of aceticacid = 4.745
C = 0.05 M
pH = 7+1/2(4.745+log0.05)
= 8.72
Explanation / Answer
When 50.0 mL of 0.050 M formic acid, HCHO2, is titrated with 0.050 M sodium hydroxide, what is the pH at the equivalence point? (Be sure to take into account the change in volume during the titration.) 9. nswer: 8.07 10. For the titration of 75.00 mL of 0.1000 M acetic acid with 0.1000 M NaOH, calculate the pH (a) before the addition of any NaOH solution, (b) after 25.00 mL of the base has been added, (c) after half of the HC2H302 has been neutralized, and (d) at the equivalence point.
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