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Chemistry / Principles of General Chemistry 2 / Chapter 4 / Problem 28P

Principles of General Chemistry | 2nd Edition | ISBN: 9780073511085 | Authors: Martin S. Silberberg

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If 38.5 mL of lead(II) nitrate solution reacts completely

Chapter 4, Problem 28P

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QUESTION:

Problem 28P

If 38.5 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.628 g of precipitate, what is the molarity of lead(II) ion in the original solution?

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QUESTION:

Problem 28P

If 38.5 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.628 g of precipitate, what is the molarity of lead(II) ion in the original solution?

ANSWER:

Problem 28P

If 38.5 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.628 g of precipitate, what is the molarity of lead(II) ion in the original solution?

Solution 28P:

Here, we are going to determine the molarity of lead ions in the original solution.

Step 1:

Molarity is defined as follows:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ solution\ in\ L}$

The balanced equation is

$Pb({NO}_{3}{)}_{2}(aq)\ +2NaI(aq)\ \rightarrow{}2Na{NO}_{3}(aq)\ +Pb{I}_{2}(s)$

Given that, the mass of $Pb{I}_{2}$ =0.628 g

The molar mass of $Pb{I}_{2}$ =461.01 g/mol

Therefore, the moles of $Pb{I}_{2}$ = (0.628 g $Pb{I}_{2}$ ) $\times{}\frac{1.0\ mol\ Pb{I}_{2}}{461.01\ \ g\ Pb{I}_{2}}$ =0.00136 mol

Thus, the moles of $Pb{I}_{2}$ is 0.00136 mol.

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