(c) F218)
. 40.00 ml of 0.11 M HCI is diluted to 100 ml with water and
then titrated with 0.1 M NaOH. The pH of the resulting
solution after addition of 10 ml titrant is
Answers
Answer:Suppose V mL of 0.1 M NaOH is used.
Number of moles of NaOH =
1000 mL/L
V mL
×0.1 mol/L=10
−4
V mol
Number of moles of HCl =
1000 mL/L
50 mL
×0.1 mol/L=0.005 mol
Out of 0.005 moles of HCl, 10
−4
V moles will be neutralised with 10
−4
V moles of NaOH.
0.005−10
−4
V moles of HCl will remain.
0.005−10
−4
V moles of HCl will give 0.005−10
−4
V moles of H
+
ions.
Total volume =(50+V) mL=
1000 mL/L
(50+V) mL
=(0.05+0.001V) L
[H
+
]=
(0.05+0.001V) L
(0.005−10
−4
V) mol
......(1)
pH=3
[H
+
]=10
−pH
=10
−3
M......(2)
But, (1) = (2)
(0.05+0.001V) L
(0.005−10
−4
V) mol
=10
−3
M
(0.05+0.001V)
(0.005−10
−4
V)
=10
−3
......(3)
Assuming V to approximately equal to 50, we can assume that the denominator 0.05+0.001V≃0.05+0.001(50)=0.05+0.05=0.1
Hence, equation (3) becomes
0.1
(0.005−10
−4
V)
=10
−3
0.005−10
−4
V=10
−4
0.005=10
−4
(1+V)
1+V=50
V=49 mL
Hence, 49 mL of 0.1 M NaOH is used.
Explanation: