Question

A sample of an alloy of silver weighing $\,0.50{\text{ }}g\,$ and containing $\,90\% \,$ silver was dissolved in conc. $\,HN{O_3}\,$ and silver was analyzed by the Volhard method. A volume of $\,25{\text{ }}ml\,$a $\,KCN\,$ solution was required for complete precipitation. The normality of $\,KCN\,$ solution is $\,(Ag = 108)\,$
A.$\,4.167\,$
B.$\,0.167\,$
C.$\,3.136\,$
D. $\,0.125\,$

Answer 1

One of the terms used to calculate the concentration of a solution is normality. It is abbreviated as $\,'N'\,$ and is also known as a solution's equivalent concentration. It is primarily used as a measure of reactive species in a solution and in circumstances involving acid-base chemistry during titration reactions.
Formula used:
$\,N = \dfrac{{mEq}}{{V\,}}\,$
Therefore $\,mEq\,$ $\, = \,$ $\,V \times N\,$
Where, $\,\,mEq\,$ is the milligram equivalent, $\,N\,$ is the normality and $\,V\,$ is the volume in millilitre ( as we are calculating milligram equivalent)
$\,mEq = \dfrac{w}{{Eq.w}}\,$
Where, $\,mEq\,$ is milligram equivalent, $\,w\,$ is weight $\,Eq.w\,$ is the equivalent weight

Complete step by step answer:
Let us analyse the given data;
We have to calculate the weight of silver. Here it is given that $\,0.50g\,$ of silver alloy contains $90\% \,$ of silver, so we can calculate weight of silver as follows;
$\,weigh{t_{Ag}} = \dfrac{{0.5 \times 90}}{{100}} = 0.45g\,$
Volume of the solution$\, = 25ml\,$
Now, let us move into the main calculation;
We know that, $\,N = \dfrac{{mEq}}{{V\,}}\,$
Therefore $\,mEq\,$ of $\,KCN = \,$ $\,V \times N\,$
Where, $\,\,mEq\,$ is the milligram equivalent, $\,N\,$ is the normality and $\,V\,$ is the volume in millilitre ( as we are calculating milligram equivalent)
We have $\,V = 25ml\,$
Number of milligrams equivalent of $KCN = N \times 25\,$
Whereas we are using a different formula to calculate number of milligram equivalent of silver as per the available data;
Number of milligrams equivalent of silver $\, = \dfrac{{Weight}}{{Equivalent\,weight}}\,$
Equivalent weight of silver is given which is $\,108g\,$and we had found out the weight to be $\,0.45g\,$
Substituting this we get;
$\,\,mE{q_{Ag}} = \dfrac{{0.45}}{{108}} \times 1000\,mg\,$(As the weights we have here is in grams, we are converting them inot milligrams)
Neutralization number of milligrams equivalent to silver is equal to the number of milligrams equivalent of $\,KCN\,$(potassium cyanide). To get normality of $\,KCN\,$ we should equalize with number of milligrams equivalent of $\,Ag\,$ with number of milligrams equivalent of $\,KCN\,$
Therefore,
$\dfrac{{0.45}}{{108}} \times 1000(mg) = N \times 25$
We get $\,N = \,\dfrac{{0.45 \times 1000}}{{105 \times 25}} = 0.1666N\,$
Hence option B is the correct answer for this question.

Additional information:
-Volhard method is the method for the determination of chlorine, bromine, and iodine in the form of halides by precipitating them with excess silver nitrate and titrating the excess with a thiocyanate solution
-Potassium cyanide is a compound with the formula KCN. This colorless crystalline salt, similar in appearance to sugar, is highly soluble in water.
-It is a crystalline salt which has no color and is highly toxic and soluble in water. It has a smell of bitter almonds and tastes like acrid with a burning sensation

Note: While normality is widely used in precipitation and redox reactions, it has some limitations. It is an uncertain measure and better choices for units are molarity or molality. For a specific chemical solution, this is not a defined value. Depending on the chemical reaction, the value will dramatically alter. To further illustrate, with different reactions, one solution will potentially contain different normalities.