Question

In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. Precipitation of solid AgCN takes place when excess Ag+ is added to the solution, above the amount needed to complete the formation of MathProcessingError. How many grams of NaCN were in the original sample?

a) 0.100 g
b) 0.250 g
c) 0.500 g
d) 1.000 g

Answer 1

To calculate the grams of NaCN in the original sample, you need to use the stoichiometry of the reaction and the volume and concentration of AgNO3 used. The correct answer is 14.09 g.

Step-by-step explanation:

To determine the grams of NaCN in the original sample, we need to use the stoichiometry of the reaction and the volume and concentration of AgNO3 used. From the balanced chemical equation:

AgNO3 + NaCN → AgCN + NaNO3

1 mole of AgNO3 reacts with 1 mole of NaCN. Therefore, the moles of NaCN in the original sample can be calculated as:

moles of NaCN = moles of AgNO3

Using the volume and concentration of AgNO3, we can calculate the moles of AgNO3:

moles of AgNO3 = volume of AgNO3 x concentration of AgNO3 = 28.72 mL x 0.0100 M = 0.2872 mol

Since the moles of NaCN and AgNO3 are equal, the grams of NaCN can be calculated using the molar mass of NaCN:

grams of NaCN = moles of NaCN x molar mass of NaCN = 0.2872 mol x 49.01 g/mol = 14.09 g

Therefore, the correct answer is d) 14.09 g.