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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

Verified Answer :

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 : 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.

\(M_1\times 25ml=(0.25M)\times (45ml)\)

\(M_1=0.45M\)

Therefore, the concentration of \(NH_3\) solution is, 0.45 M

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