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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