\"When a strong base is gradually added drop wise to a weak acid, the pH changes
ID: 1060997 • Letter: #
Question
"When a strong base is gradually added drop wise to a weak acid, the pH changes at each addition. When the appropriate quantity of base has been added to react with all of the acid, the pH changes sharply, indicating the endpoint of the titration. A plot of pH versus volume of base added gives what is known as a titration curve" Consider the titration of 25.00 mL of 0.1000 M benzoic acid with 0.1000 M NaOH. Write a balanced chemical equation for the titration reaction using C6H5COOH for benzoic acid's formula. Determine the volume of NaOH solution required to reach the endpoint. The chart on the next page has entries for several steps along the titration curve. To calculate the pH at each step, you first must understand the chemistry at that step, and then you can decide the appropriate method to calculate the pH. For each volume listed, (i) list the major species in solution, (ii) determine whether the Ka equation. Kb equation, buffer equation, or solution equilibrium equation is appropriate for the calculation of the solution [H30+] and pH. and (iii) complete the calculations.Explanation / Answer
a) balanced equation :
C5H5COOH + NaOH ---------------------> C6H5COONa + H2O
b)
at equivalence point : millimoles of acid = millimoles of base
25 x 0.1 = V x 0.1
V = 25 mL
NaOH volume needed = 25 mL
c)
pKa = 4.2
millimoles of benzoic acid = 25 x 0.1 = 2.5
a) 0 ml NaOH added
pH = 1/2 (pKa + log C)= 1/2 (4.2 + log (0.1 ) = 2.60
pH = 2.60
b) 5 ml NaOH
NaOH millimoles = 5 x 0.1 = 0.5
C5H5COOH + NaOH ---------------------> C6H5COONa + H2O
2.5 0.5 0 0 --------------------initial
2.0 0 0.5 0.5
pH = pKa + log[salt/acid]
= 4.20 + log (0.5/2.0)
pH= 3.60
c) 12.5ml NaOH
NaOH millimoles = 12.5 x 0.1 = 1.25
it is half - equivalence point . so here
pH = pKa
pH = 4.20
d) 20ml NaOH
NaOH millimoles = 20 x 0.1 = 2
C5H5COOH + NaOH ---------------------> C6H5COONa + H2O
2.5 2 0 0
0.5 0 2 2
pH = 4.2 + log [2 / 0.5]
pH = 4.80
e) 25 ml NaOH added :
it is equivalence point salt only remains :
so concentration of salt = C = 2.5 / 50 = 0.05 M
pH = 7 + 1/2 (pKa + log C)
= 7 + 1/2 (4.20 + log 0.05)
= 8.45
pH = 8.45
f) 30 ml NaOH
NaOH millimoles = 30 x0.1 = 3
C5H5COOH + NaOH ---------------------> C6H5COONa + H2O
2.5 3 0 0
0 0.5 2.5
solution only contain strong base NaOH so pH decide by base only
[OH] = 0.5 /total volume
= 0.5 / (30+ 25)
= 0.00909 M
pOH = 2.04
pH = 11.96
pH = -log [H3O+]
pH= 12.21
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