🙂 For Qustion like this one you must now the concept, let me explain a little about that:
first: Hcl and NaoH both are strong acid and base (k is very big number)
second: they are monoporetic and mono hydroxide(actualy make it easier) that means, one H+ and one oH-naturalize each other.
third(very important point): for both Molarity and Normality are equal(because the second point that I alredy mentioned) in other hand Mollar is 1, means Normality is 1 and if not , you must calculate the normality( recall from acid and base soultion ===> N= molarity x n)
-(PH)
forth: concept of PH for acid ===> N=[H+]= 10
now ve can attack the problom:
for NaoH : V1=20 ml
Cm(molarity)=0.1 ====> N1=0.1
for Hcl : V2=?
Cm=0.1 ==========> N2=0.1
fina soultion after adding Acid and Base: -2
PH= 2 ====> N(total)=10 = 0.01
V(total)= V1 + V2= 20 + V2
now, we can go for formula with knowing this concept that :we added acid and nutralized the base (N1V1=N2V2) and then add more acid and getting acidity solution(PH=2 means acidity solution) in other hand for sure we must add 20 cc acid and adding alittle more to get acidity solution.
now we have formula
N2V2-N1V1= NV=====> (0.1 x V2 )- (0.1 x 20 )= 0.01 x(20+ V2)
beacause V total = V1+V2
now find the V2 from above you will get this, in the end: V2=(20.2/0.99)
and the answer wil be : 20.4 ml 🙂!🙂
good luck for ur DAT
