Question

25. 0 ml of a 0. 100 m solution of NH3 is titrated with 0. 250m HCl. After 10. 0 ml of the hcl has been added, the resultant solution is?

Answer 1

Main Answer: The resultant solution will have 0.125M of HCl.

Explanation:

Given:

Volume of NH3  V1= 25ml = 0.025lit

Molar concentration of NH3  M1= 0.1M

It is titrated with HCl

Molar concentration of HCl M2= 0.25M

Volume of HCl  V2= ?ml

At equilibrium,

M1V1 = M2V2

V2 = M1V1/M2

V2 = (0.1*0.025)/0.25

V2= 0.01lit =0.01*1000 = 10ml

Number of moles = M2*V2 = 0.25 * 0.01 = 0.0025

Now 10ml of HCl is added.

then,

V2 = 10 + 10 = 20ml

Molar concentration of HCl = number of moles/Volume

                                               = 0.0025/0.02

                                               = 0.125M

The resultant solution will have 0.125M of HCl.

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