100 mL solution of 0.1 N HCl was titrated with 0.2 N NaOH solutions. The titration was discontinued afteradding 30 mL of NaOH solution. The remaining titration was completed by adding 0.25 N KOH solution. Thevolume of KOH required from completing the titration is:

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70 mL

35 mL

32 mL

16 mL

answer is D.

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Detailed Solution

Total mEq of HCI= 100 × 0.1 = 10 Total mEq of NaOH = 30 × 0.2=6 mEq of HCl left = 10-6=4 mEq of HCl = mEq of KOH 4=0.25 ×V ; V⇒16mL

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