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.

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

$M_1=0.45M$

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