Question

200ml of 10N Hcl was mixed with 100 ml of 15N Naoh solutions find out the new normality of the resultant solution​

Answer 1

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Answer:

0.05 N

Explanation:

Given Normality of HCI = 10N,

Volume of HCl = 200 ml

Normality of NaOH = 15N,

Volume of NaOH = 100 ml

∴ Miliequivalents of HCl =10×200=2000

∴ Miliequivalents of NaOH =15×100=150

HCl+NaOH→NaCl+H

Miliquivalents before reaction 10 20 0 0

Milliequivalents after reaction 0 10 10 10

No. of moles left in solution = 20 - 10 = 10

Normality of resulting solution = No. of moles left in solution/Total volume of solution

= 10/100+100

= 10/200

=0.05N

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