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A sample of pure As2O3 (m.m. 197.84) weighing 0.2068 g is dissolved in NaOH and then acidified with HCI. The resulting solution is then titrated with 42.45 ml of the KMnO4 solution. What is the molarity of the KMnO4 solution? hзaso3 + H2O→ HзASO4 + 2h + 2e a) 0.09850 b) 0.04925 c) 0.01970 d) 0.03254
Accepted Answer
The balanced equation for the reaction is:
5HзASO3 + 2KMnO4 + 6HCl → 5HзASO4 + 2KCl + 2MnCl2 + 3H2O
From the balanced equation, we can see that 2 moles of KMnO4 react with 5 moles of HзASO3.
First, calculate the moles of As2O3:
0.2068 g As2O3 / 197.84 g/mol = 0.001045 mol As2O3
Since each mole of As2O3 produces 2 moles of HзASO3:
0.001045 mol As2O3
2 = 0.002090 mol HзASO3
Now, calculate the moles of KMnO4 used in the titration:
0.002090 mol HзASO3
(2 mol KMnO4 / 5 mol HзASO3) = 0.000836 mol KMnO4
Finally, calculate the molarity of the KMnO4 solution:
0.000836 mol KMnO4 / 0.04245 L = 0.01970 M
Therefore, the molarity of the KMnO4 solution is 0.01970 M (C).
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