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What are the concentrations of $\mathrm{Ag}^{+}, …
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Problem 79 Hard Difficulty

In a titration of cyanide ion, 28.72 $\mathrm{mL}$ of 0.0100 $\mathrm{M} \mathrm{AgNO}_{3}$ is added before precipitation begins. [The reaction of $\mathrm{Ag}^{+}$ with $\mathrm{CN}^{-}$ goes to completion, producing the $\mathrm{Ag}(\mathrm{CN})_{2}^{-}$ complex.] Precipitation of solid AgCN takes place when excess Ag^+ is added to the solution, above the amount needed to complete the formation of $\operatorname{Ag}(\mathrm{CN})_{2}-$ How many grams of NaCN were in the original sample?

Answer

0.0281 $\mathrm{g}$

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Chapter 15

Equilibria of Other Reaction Classes

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Video Transcript

The reaction involved in formation of silver sinite will be e g, positive, plus 2 c n negative, which gives us a g c n 2 negative. The tritration of sodium sinite with sulphur nitride involves the formation of a g c after the whole. Sinite irons are exhausted. Silver, positive iron, furnished by excess of a g n o 3, can be the act, so the reaction is represented as a g c, n, 2 negative plus a g positive gives 2 a g c n, as the precipitation begins after 28.72 ml. So let me write it as the precipitation begins. After 28.72 ml of 0.0100 molar, a g n o 3, we need to calculate the number of moles of silver ion present and of sin will be equal to molarity multiplied with volume substituting in those values. Molarity is 0.0100 and volume will be 28.72 divided by 1000 l per liter. This will be so. The number of moles turn out to be 2.87 multiplied 10 raised to the power negative 4 mole. As for the reaction, the formation of silver sinite moles of sinite involved will be double as compared to silver, positive iron, so calculating this and c negative will be equal to 2 times the mole of a g positive, which will be equal to 5.7410 raised to the Power negative 4 moles.

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