Question

Concentrated HCl solution is 37.0% HCl and has a density of 1.19 g/mL. A dilute solution of HCl is prepared by diluting 4.50mL of this concentrated HCl solution to 100mL with water. Then 10mL of this dilute HCl solution reacts with an $AgN{O}_3$ solution. Calculate the volume of 0.108M $AgN{O}_3$ solution required to precipitate all the chloride as $AgC{l}_{(s)}$.

Answer 1

Hint: The formula to find molarity and molarity of dilute solution obtained from concentrated solution are as below.
$$M={\displaystyle \frac{\text{Weight of solute}\times \text{1000}}{\text{Molecular weight of solute}\times \text{Volume of the solute(mL)}}}$$
$$M_1{V}_1=M_2{V}_2$$

Complete Step-by-Step Solution:
We will first calculate the molarity of concentrated HCl solution. From that, we will find the molarity of dilute HCl solution and finally from it, we will find the volume of $AgN{O}_3$ solution required to complete precipitation.
- We are given that the concentrated HCl solution is 37%. So, we can say that if the solution is of 100 g, then the weight of dissolved HCl will be 37g.
- Density of the solution is given 1.19 g/mL. Now, we will use the formula of the density given below.
$$\text{Density = }{\displaystyle \frac{\text{Weight}}{\text{Volume}}}$$
So, $1.19={\displaystyle \frac{100}{\text{Volume}}}$
Thus, we can write that Volume = ${\displaystyle \frac{100}{1.19}}=84.03mL$
We know that Molecular weight of HCl = Atomic weight of H + Atomic weight of Cl
So, Molecular weight of HCl = 1 + 35.5 = 36.5 g/mol
Now, we will find the molarity of the solution.
$$M={\displaystyle \frac{\text{Weight of solute}\times \text{1000}}{\text{Molecular weight of solute}\times \text{Volume of the solute(mL)}}}$$
So,
$$M={\displaystyle \frac{37\times 1000}{36.5\times 84.03}}=12.06M$$
Now, we will find the molarity of diluted solution of HCl using the equation given below.
$$M_1{V}_1=M_2{V}_2$$
where $M_1$ = Molarity of con. HCl solution = 12.06M
$V_1$ = volume of cone. HCl solution taken to make dilute solution = 4.50 mL
$M_2$ =Molarity of diluted solution
$V_2$ =Volume of diluted solution of HCl = 100mL
So, we can write that
$$12.06\times 4.50=M_2\times 100$$
So,
$$M_2={\displaystyle \frac{12.06\times 4.50}{100}}=0.5427M$$
Now, we will take a look at the reaction between HCl and $AgN{O}_3$.
$$HCl+AgN{O}_3\to AgCl+HN{O}_3$$
 - Now, we are given that 10mL of this 0.5427M solution is allowed to react with 0.108M $AgN{O}_3$ solution.
- From the definition of molarity, we can say that 1000mL of 0.5427M solution will contain 0.5427 moles of HCl. So, 10mL of 0.5427M solution will contain 0.005427 moles.
- Now, we will require 0.005427 moles of $AgN{O}_3$ to have complete precipitation.
- From the definition of molarity, we can say that 0.108 moles of $AgN{O}_3$ will be obtained from 1000mL of 0.108M $AgN{O}_3$ solution. So, 0.005427 moles of $AgN{O}_3$ can be obtained from ${\displaystyle \frac{0.005427\times 1000}{0.108}}=50.25mL$

Thus, we found that we will require 50.25mL of 0.108M $AgN{O}_3$ solution to ensure complete precipitation.

Note: Remember that whenever we make a solution from another solution of known concentration, then we can use the formula $M_1{V}_1=M_2{V}_2$ to find the volume or molarity whichever is unknown. If we are given normality, then we can use normality of both the solutions in the same formula.