Question

what amount of water is added in 40 ml of NaOh(0.1N) which is neutralized by 50 ml of HCL(0.2)

Answer 1

Number of gram equivalents(eq) = moles x acidity of a base (or bascity of an acid.)
Or
= Molarity x Volume in litres x acidity of a base (or bascity of an acid)
So,
Eq(NaOH) = 50/1000 x 0.1 x 1 = 0.005 eq
Eq(CH3COOH) = 20/1000 x0.25 x1 = 0.005 eq
That means acid = base. Hence a salt would be formed in the solution with total volume = 50 + 20 = 70 ml
The pH of a strong base weak acid salt solution =
7 + 0.5(pKa +logC) where C is concentration of salt
concentration of salt = eq/volume =1000 x 0.005/70
= 0.0714
So
pH = 7 + 0.5(4.74 +log0.0714)
pH = 8.8
no of equivalents of NaOH= no of euivalents of HCl
(40+x)0.1=50*0.2
100=40+x
x=60

not confirmed