calculate the pH of mixture of 10ml of 0.1m h2so4 + 10ml of 0.1m koh
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1
Answer:
the answer is 1.3.
Explanation:
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.
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