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10 mL of a solution of NaOH is found to be completely neutralized 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 neutralize it will be :
(a) 4 ml (b) 8 ml
(c) 12 ml (d) 16 ml
d) 16 ml
Explanation: HCL reacts with NaOH to form NaCl + H2O. 8 mL of HCL (hydrochloric acid) solution completely neutralises 10 mL of NaOH (sodium hydroxide). The amount of NaOH solution has been doubled in the questions; thus, the amount required of the HCL solution will also be doubled. 16 mL of HCL (hydrochloric acid) solution completely neutralises 20 mL of NaOH (sodium hydroxide).