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10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a given solution of HCl. If we take 20 mL of the same solution of NaOH, the amount HCl solution (the same solution as before) required to neutralise it will be:

(a) 4 mL           (b) 8 mL         (c) 12 mL                 (d) 16 mL



(d.) 16 mL


Explanation :

The ratio of acid to base in a neutralisation reaction always remains constant for a pair of acid and base. In this case 10mL of NaOH solution is being neutralised by 8 mL of HCl solution. We now doubled the amount of NaOH solution to 20 mL . Hence the amount of HCl solution required to neutralise the 20 mL of NaOH will also be doubled that is 8x2 which is 16 mL. 

Therefore 16 mL of the same HCl solution will neutralise 20 mL of NaOH solution and d is the correct option.

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Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 13 years and a teacher from the past 17 years. He teaches Science, Economics, Accounting and English at Teachoo