NCERT Solution for Class Tenth Science Chapter 2
Acids, Bases and Salts
Exercises Questions
Page No-18
Questions No-3
(a) 4 mL (b) 8 mL (c) 12 mL (d) 16 mL
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
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(d) 16 ml
Because for the neutralisation, the ratio of the acid and base should be same.
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(d) 16 mL of HCl solution will be required to neutralise.
To solve this problem, you can use the concept of equivalent concentrations. In a neutralization reaction, the moles of acid (HCl) are equal to the moles of base (NaOH) when they react in a 1:1 ratio. Given that 10 mL of NaOH solution neutralizes 8 mL of HCl solution, it means the two solutions have equivalent concentrations.
Now, if you have 20 mL of the same NaOH solution, you can find the amount of HCl solution needed to neutralize it. Since the concentrations are equivalent, you can set up the following proportion:
(10 mL of NaOH) / (8 mL of HCl) = (20 mL of NaOH) / (x mL of HCl)
Now, you can solve for x:
x = (20 mL of NaOH * 8 mL of HCl) / 10 mL of NaOH
x = 160 mL of HCl
So, 160 mL of the same HCl solution will be required to neutralize 20 mL of the NaOH solution.