TS EAMCET · Chemistry · Redox Reactions
The concentration of oxalic acid is ' \(x\) ' \(\mathrm{mol} \mathrm{L}^{-1}\). \(40 \mathrm{~mL}\) of this solution reacts with \(16 \mathrm{~mL}\) of \(0.05 \mathrm{M}\) acidified \(\mathrm{KMnO}_4\). What is the \(\mathrm{pH}\) of ' \(x\) ' \(\mathrm{M}\) oxalic acid solution ? (Assume that oxalic acid dissociates completely)
- A 1.3
- B 1.699
- C 1
- D 2
Answer & Solution
Correct Answer
(A) 1.3
Step-by-step Solution
Detailed explanation
Oxalic acid \(=x \mathrm{~mol} / \mathrm{L}\) Oxalic acid \(\mathrm{KMnO}_4\) \(M_1 V_1=M_2 V_2\) \(40 \mathrm{~mL} \times x=16 \mathrm{~mL} \times 0.05\) \(x=\frac{16 \times 0.05}{40}=\frac{1}{50}\) \(x=\frac{1}{50} \mathrm{M}\) Now convert molarity into normality \(N \times\)…
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