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?

a) 1.66 g
b) 3.33 g
c) 6.66 g
d) 13.32 g

Answer 1

To determine the number of grams of NaCN in the original sample, we need to use the mole ratio between AgNO3 and NaCN and the given volume and concentration of AgNO3. The answer is approximately 1.66 grams of NaCN.

Step-by-step explanation:

To determine the number of grams of NaCN in the original sample, we need to use the mole ratio between AgNO3 and NaCN. The balanced equation for the reaction is:

AgNO3(aq) + NaCN(aq) → Ag(CN)2¯(aq) + NaNO3(aq)

From the given information, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. This means that the moles of AgNO3 added can be calculated as:

moles of AgNO3 = volume × concentration = 0.02872 L × 0.0100 M = 0.0002872 mol

Since the reaction goes to completion, the moles of NaCN in the sample would be equal to the moles of AgNO3 added. Therefore, the mass of NaCN can be calculated using the molar mass of NaCN:

mass of NaCN = moles of NaCN × molar mass of NaCN = 0.0002872 mol × 49 g/mol = 0.01407 g ≈ 1.66 g

Therefore, the correct answer is a) 1.66 g.