Answer to Question #91281 in General Chemistry for Morgan Washington

Question

what volume of 0.125 M HCL is needed to titrate 50.0 ML of 0.0500 M NH3 solution. ML

Answer 1

$HCl(aq) + NH_3 (aq) = NH_4Cl(aq)$

At the equivalence point, the number of moles of HCl added is equal to the initial number of moles of NH3, because the analyte is completely neutralized.

$n(NH_3)=c\times V = 0.0500 \frac{mol}{L} \times 50.0\times 10^{-3} L = 0.0025 mol$

$n(NH_3) = n(HCl)$

$n(HCl) = 0.0025 mol$

$c = \frac{n}{V}$

$\therefore V = \frac{n}{c} = \frac {0.0025 mol}{0.125\frac{mol}{L}}= 0.02 L = 20.0 mL$