Question
Question: 22. 1 litre solution of unknown molarity is titrated by taking its 50 mL solution against KI solutio...
- 1 litre solution of unknown molarity is titrated by taking its 50 mL solution against KI solution is strong acid medium of excess HCl. The equivalence point was detected when 10 mL of 0.1 M KI was consumed. The molarity of KIO3 solution is :
A
4 × 10–4 M
B
2 × 10–2 M
C
4 × 10–3 M
D
2 × 10–3 M
Answer
4 × 10–3 M
Explanation
Solution
To find the molarity of KIO3, we first calculate the moles of KI used in the titration. The volume of KI used is 10 mL of 0.1 M solution, which gives us 0.001 moles of KI. Since the reaction between KIO3 and KI is a 1:5 ratio, the moles of KIO3 will be 0.001 moles / 5 = 0.0002 moles. The molarity of KIO3 in the original 1 L solution is then 0.0002 moles / 0.05 L = 0.004 M, which is 4 × 10–3 M.