Question

A 25 ml of solution of 0.5 M Naoh is titrated until neutralized into a 50 ml sample of hcl , what is the concentration of hcl?

Answer 1

Answer:

A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was the concentration of the HCl? Every mole of NaOH will have one mole of OH-. Therefore [OH-] = 0.5 M.

Answer 2

Given:

A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl.

To Find:

The concentration of HCl.

Solution:

To find the concentration of HCl we will follow the following steps:

As we know,

Molarity is a number of moles of solute present in 1lite of solution.

And,

An equal number of grams equivalent of acid can neutralise a base in a 1 lite solution.

N1V1 = N2V2

Normality = molarity × n factor

n1M1V1 = n2M2V2

the factor is same and equal to 1 for both NaOH and HCl.

M1V1 = M2V2

Here,

M1 = molarity of a solution of NaOH

M2 = molarity of a solution of HCl

V1 = volume of NaOH used

V2 = volume of HCl used

n1 and n2 is n factor of NaOH and HCl.

Because NaOH = Na+ + OH- and HCl = H+ + Cl-

1 electron loss or gain so the n factor is 1.

Now,

According to the question by putting values in the above equation we get,

25 × 0.5 = M2 × 50

M2 = 4 M

Henceforth, the concentration of HCl is 4M.