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.
\[M = \dfrac{{{\text{Weight of solute}} \times {\text{1000}}}}{{{\text{Molecular weight of solute}} \times {\text{Volume of the solute(mL)}}}}\]
\[{M_1}{V_1} = {M_2}{V_2}\]

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.
\[{\text{Density = }}\dfrac{{{\text{Weight}}}}{{{\text{Volume}}}}\]
So, $1.19 = \dfrac{{100}}{{{\text{Volume}}}}$
Thus, we can write that Volume = $\dfrac{{100}}{{1.19}} = 84.03mL$
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.
\[M = \dfrac{{{\text{Weight of solute}} \times {\text{1000}}}}{{{\text{Molecular weight of solute}} \times {\text{Volume of the solute(mL)}}}}\]
So,
\[M = \dfrac{{37 \times 1000}}{{36.5 \times 84.03}} = 12.06M\]
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,
\[{M_2} = \dfrac{{12.06 \times 4.50}}{{100}} = 0.5427M\]
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 $\dfrac{{0.005427 \times 1000}}{{0.108}} = 50.25mL$

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.