0.1N solution of a diabasic acid can before the better by dissolving 0.45 gram of a sudden water and editing 100 ml.The Molarity of nitrate ions the resulting mixture will be
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According to the given Question,the reaction is
2HNO3+2H2O <---------------> 2HNO2+2H2O+O2
Given Normality of a dibasic acid =0.1N
Therefore Molarity = Normality/acidity
Acidity=2,,,,As it is a dibasic acid
Molarity of NO3 = 0.1/2 = 0.05M
Molarity of OH= 0.45/18 * 1000/100
by solving we get , Molarity of OH = 0.25M
Volume of NO3 = n/Molarity of NO3
n=2 (beacuse ,as per the reaction number of nitrate ions are 2)
Volume of NO3= 2/0.05
by solving we get ,V=40 Litre = 40000 ml
Therefore ,, Molarity of resulting mixture = M1V1+M2V2 /V1+V2
M= 0.05*40000+0.25*100 / 40000+100
=5*400+25 / 40100
=2000+25/40100
= 2025/40100
=0.05M
Molarity of the resultant solution=0.05M
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