Question

If 45.0 mL of 0.25 M HCl is required to completely neutralize 25.0 mL of NH3, what is the concentration of the NH3 solution? Show all of the work needed to solve this problem.

HCl + NH3 yields NH4Cl

Answer 1

hcl + nh3 -> nh4cl (balanced eqn)

no. of mol of hcl = vol. (L) x molarity = 0.045 × 0.25 = 0.01125mol

ratio of hcl:nh3 after balancing eqn = 1:1

no. of mol of nh3 that is completely neutralised by hcl = 0.01125 × 1 = 0.01125mol

therefore, concentration of nh3 = mol / total volume (L) = 0.01125mol / 0.025L= 0.45M

Answer 2

Answer : The concentration of
`NH_3` solution is, 0.45 M

Solution :

The given balanced reaction is,

`HCl+NH_3\rightarrow NH_4Cl`

The moles ratio of
`HCl` and
`NH_3` is, 1 : 1 that means 1 mole of HCl neutralizes by the 1 mole of ammonia.

According to the neutralization law,

`M_1V_1=M_2V_2`

where,

`M_1` = molarity or concentration of
`NH_3` solution = ?

`V_1` = volume of
`NH_3` solution = 25 ml

`M_2` = molarity of concentration HCl solution = 0.25 M

`V_2` = volume of HCl solution = 45 ml

Now put all the given values in the above law, we get the concentration of
`NH_3` solution.

`M_1* 25ml=(0.25M)* (45ml)`

`M_1=0.45M`

Therefore, the concentration of
`NH_3` solution is, 0.45 M