Question #b8733

Question

1300 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-24T05:45:30

In case of acid base neutralization we know the relation

#S_axxV_a=S_bxxV_b#

Where

#S_a-> "strength of acid solution in Normality" =?#

#V_a->"Volume of acid solution"=2mL#

#S_b-> "strength of base solution in Normality"=N/5 #

#V_b->"Volume of acid solution"=10mL#

Hence

#S_axxV_a=S_bxxV_b#

#=>S_axx2=N/5xx10#

#=>S_a=1(N)#