Question

Upon mixing 50 mL of 0.1 M lead nitrate solution with 50 mL of 0.05 M chromic sulphate solution, precipitation of lead sulphate solution takes place. How many moles of lead sulphate are formed and what is the molar concentration of chromic sulphate left in the solution?
a.) 0.005, 0.0084
b.) 0.0084, 0.005
c.) 0.005, 0.00084
d.) 0.05, 0.00084

Answer 1

Hint: To solve this question, write the balanced equation for the reaction. Find the limiting reagent. Then, calculate the number of moles for the individual reactant and product using the data given in the question.

Complete step by step answer:
- Let us start by writing the equation for the reaction given in the question –
\[3Pb{{(N{{O}_{3}})}_{2}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\to 3PbS{{O}_{4}}\downarrow +2Cr(N{{O}_{3}})\]
As we can see, 3 moles of lead nitrate reacts with 1 mole of chromic sulphate to give 3 moles of lead sulphate and 2 moles of chromium nitrate.
Given,
Lead nitrate = 50 mL of 0.1 M = \[50\text{ }\times \text{ }0.1\text{ }mmol\text{ }=\text{ }5\text{ }mmol\]
Chromic sulphate = 50 mL of 0.05 M = \[50\text{ }\times 0.05\text{ }mmol\text{ }=\text{ }2.5\text{ }mmol\]
Mmol is known as millimoles, defined as the amount of substance equal to a thousand of a mole.
Therefore, we can say that –
3 mmol of $Pb{{(N{{O}_{3}})}_{2}}$ will react with 1 mmol of $C{{r}_{2}}{{(S{{O}_{4}})}_{3}}$ .
1 mmol of $Pb{{(N{{O}_{3}})}_{2}}$ will react with \[\dfrac{1}{3}\] mmol of $C{{r}_{2}}{{(S{{O}_{4}})}_{3}}$.
5 mmol of $Pb{{(N{{O}_{3}})}_{2}}$ will react with \[\dfrac{1}{3}\times \text{5}\] mmol of $C{{r}_{2}}{{(S{{O}_{4}})}_{3}}$ = 1.66 mmol of $C{{r}_{2}}{{(S{{O}_{4}})}_{3}}$.
So, $Pb{{(N{{O}_{3}})}_{2}}$ is a limiting reagent.
Since, 3 mmol of $Pb{{(N{{O}_{3}})}_{2}}$ will give 3 mmol of $PbS{{O}_{4}}$ ,
Therefore, 5 mmol of $Pb{{(N{{O}_{3}})}_{2}}$ will give 5 mmol of $PbS{{O}_{4}}$.
Converting millimoles to moles,
\[5\text{ }mmol\text{ }=\text{ }\dfrac{5}{1000}\text{ }mol\text{ }=\text{ }0.005\text{ }mol\]
- Now, as we know lead nitrate is the limiting reagent. Therefore, some amount of chromic sulphate will remain in the solution.
- From the above calculations, we know 1.66 mmol is used in the reaction.
The remaining mmol = \[2.5\text{ }\text{ }1.66\text{ }mmol\text{ }=\text{ }0.84\text{ }mmol\]
The solution contains 50 mL lead nitrate and 50 mL chromium sulphate.
Total volume = \[50\text{ }+\text{ }50\text{ }=\text{ }100\text{ }mL\]
Molar Concentration = \[\dfrac{0.84}{100}\text{ }=\text{ }0.0084\text{ }M\]
Therefore, the answer is 0.005 moles of lead sulphate are formed. Molar concentration of chromic sulphate left in the solution is 0.0084 M.
So the correct answer is “A”:

Note: In this question, do not confuse moles with molar concentration. Molar concentration refers to the molarity. Molarity of any solution is defined as, “number of moles of solute per litre of solution”.