Chapter 4, Problem 4.131QP

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for reaction of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ should be written and molar concentration of ${\text{Pb}}^{\text{2+ }}$ in given polluted water sample should be calculated.

Concept introduction:

Net ionic equation:

Only the species, which are contributing in participate, are written the form of ions reaction is called net ionic equation.

Precipitation reaction:

If precipitate is formed, when two solutions mixed together is called precipitation reaction.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

(b)

Interpretation Introduction

Interpretation:

The net ionic equation for reaction of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ should be written and molar concentration of ${\text{Pb}}^{\text{2+ }}$ in given polluted water sample should be calculated.

Concept introduction:

Net ionic equation:

Only the species, which are contributing in participate, are written the form of ions reaction is called net ionic equation.

Precipitation reaction:

If precipitate is formed, when two solutions mixed together is called precipitation reaction.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$