Question: In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO₃ is added before precipitation begins. [The reaction of Ag⁺ with CN^– goes to completion, producing the Ag(CN)₂ ⁻ complex.] Precipitation of solid AgCN takes place when excess Ag+ is added to the solution, above the amount needed to complete the formation of Ag(CN)₂ ⁻. How many grams of NaCN were in the original sample?

The reaction

\({\rm{A}}{{\rm{g}}^ + } + 2{\rm{C}}{{\rm{N}}^ - } \to {\rm{Ag}}({\rm{CN}})_2^ - \)

Let us calculate the mass of NaCN that was in the original sample.

Calculate the number of moles of AgNO₃ is

\(\begin{array}{*{20}{c}}{{n_{{\rm{AgN}}{{\rm{O}}_3}}} = 0.0100{\rm{M}} \cdot 28.72{\rm{mL}} \cdot \frac{{1{\rm{L}}}}{{1000{\rm{mL}}}}}\\{ = 2.872 \cdot {{10}^{ - 4}}{\rm{mol}}}\end{array}\)

Since 1 mole of AgNO₃ dissociates completely into 1 mole of Ag⁺ and 1 mole of NO₃^- , the number of moles of Ag⁺is\(2.872 \cdot {10^{ - 4}}{\rm{mol}}\).

Since 1 mole of Ag⁺reacts with 2 moles of CN^-, the number of moles of CN^- is

\(\begin{array}{*{20}{c}}{{n_{C{N^ - }}} = 2 \cdot {n_{{\rm{A}}{{\rm{g}}^ + }}}}\\{ = 2 \cdot 2.872 \cdot {{10}^{ - 4}}{\rm{mol}}}\\{ = 5.744 \cdot {{10}^{ - 4}}{\rm{mol}}}\end{array}\)

Therefore, the number of moles of NaCN is\(5.744 \cdot {10^{ - 4}}{\rm{mol}}\).

\(\begin{array}{*{20}{c}}{{m_{NaCN}} = {n_{NaCN}} \cdot {M_{NaCN}}}\\{ = 5.744 \cdot {{10}^{ - 4}}{\rm{mol}} \cdot 49.0072{\rm{g}}/{\rm{mol}}}\\{ = 2.815 \cdot {{10}^{ - 2}}{\rm{g}}}\end{array}\)