20 g of NaOH is dissolved in 200 mL of
water. 10 mL of this solution is diluted to
500 mL by water. This resultant solution is
mixed with 500 mL of 0.1 M HCl solution.
Find pH of the final mixture ? (log 5=0.7)
Answers
Answer:
(a) Total number of moles present in 10 mL of 0.2 M calcium hydroxide are
1000
10×0.2
=0.002 moles.
Total number of moles present in 25 mL of 0.1 M HCl are
1000
25×2
=0.0025 moles.
Ca(OH)
2
+2HCl→CaCl
2
+2H
2
O
1 mole of calcium hydroxide reacts with 2 moels of HCl.
0.0025 moles of HCl will react with 0.00125 moles of calium hydroxide.
Total number of moles of calcium hydroxide unreacted are 0.002−0.00125=0.00075 moles.
Total volume of the solution is 10+25=35 mL.
The molarity of the solution is
35
0.00075×1000
=0.0214M
[OH
−
]=2×0.0214=0.0428M
pOH=−log0.0428=1.368
pH=14−pOH=14−1.368=12.635
(b) Total number of moles present in 10 mL of 0.01 M H
2
SO
4
is
1000
10×0.01
=0.0001 mol.
Total number of moles present in 10 mL of 0.01 M Ca(OH)
2
=
1000
10×0.01
=0.0001 mole.
Ca(OH)
2
+H
2
SO
4
→CaSO
4
+2H
2
O
0.0001 mole of Ca(OH)
2
will react completely with 0.0001 mole of H
2
SO
4
.
Hence, the resuting solution is neutral with pH 7.0
(c) Total number of moles in 10 mL of 0.1 M H
2
SO
4
=
1000
10×0.1
=0.001 mole.
Total number of moles present in 10 mL of 0.1 M KOH =
1000
10×0.1
=0.001 mole.
2KOH+H
2
SO
4
→K
2
SO
4
+2H
2
O
2 moles of KOH reacts with 1 mole of H
2
SO
4
.
0.001 mole of KOH will react with 0.0005 mole of H
2
SO
4
.
Number of moles of H
2
SO
4
left =0.001−0.0005=0.005M
Volume of solution is 10+10=20mL
Molarity of the solution is
20
0.0005×1000
=0.025M.
[H
+
]=2×2.5×10
−2
=0.05M
pH=−log[H
+
]=−log0.05=1.3.