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To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorus acid `(H_(3)PO_(3))`, the volume of 0.1 M aqueous KOH solution required is :A. 10 mLB. 20 mLC. 40 mLD. 60 mL |
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Answer» Correct Answer - C `H_(3)PO_(3)+2KOH to K_(2)HPO_(3)+2H_(2)O` `(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))` `(0.1xx20)/(1)=(0.1xx V_(2))/(2)` `V_(2)= 40 mL` |
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