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Volume of a diluted solution

What volume of 0.12 M HBr can be prepared from 11 mL (0.011 L) of 0.45 M HBr?

Solution

We are given the volume and concentration of a stock solution, V 1 and C 1 , and the concentration of the resultant diluted solution, C 2 . We need to find the volume of the diluted solution, V 2 . We thus rearrange the dilution equation in order to isolate V 2 :

C 1 V 1 = C 2 V 2 V 2 = C 1 V 1 C 2

Since the diluted concentration (0.12 M ) is slightly more than one-fourth the original concentration (0.45 M ), we would expect the volume of the diluted solution to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields:

V 2 = ( 0.45 M ) ( 0.011 L ) ( 0.12 M ) V 2 = 0.041 L

The volume of the 0.12- M solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate.

Check your learning

A laboratory experiment calls for 0.125 M HNO 3 . What volume of 0.125 M HNO 3 can be prepared from 0.250 L of 1.88 M HNO 3 ?

Answer:

3.76 L

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Volume of a concentrated solution needed for dilution

What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH?

Solution

We are given the concentration of a stock solution, C 1 , and the volume and concentration of the resultant diluted solution, V 2 and C 2 . We need to find the volume of the stock solution, V 1 . We thus rearrange the dilution equation in order to isolate V 1 :

C 1 V 1 = C 2 V 2 V 1 = C 2 V 2 C 1

Since the concentration of the diluted solution 0.100 M is roughly one-sixteenth that of the stock solution (1.59 M ), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:

V 1 = ( 0.100 M ) ( 5.00 L ) 1.59 M V 1 = 0.314 L

Thus, we would need 0.314 L of the 1.59- M solution to prepare the desired solution. This result is consistent with our rough estimate.

Check your learning

What volume of a 0.575- M solution of glucose, C 6 H 12 O 6 , can be prepared from 50.00 mL of a 3.00- M glucose solution?

Answer:

0.261 L

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Key concepts and summary

Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.

Chemistry end of chapter exercises

Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.

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What information do we need to calculate the molarity of a sulfuric acid solution?

We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.

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Practice Key Terms 9

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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