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2.4 Dilutions of Standard Solutions

  • Page ID
    32264
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    Imagine we have a salt water solution with a certain concentration. That means we have a certain amount of salt (a certain mass or a certain number of moles) dissolved in a certain volume of solution. Next we willl dilute this solution - we do that by adding more water, not more salt:

    before dilution after dilution

    Before Dilution After Dilution

    The molarity of solution 1 is

    \[ M_1 = \dfrac{\text{moles}_1}{\text{liter}_1}\]

    and the molarity of solution 2 is

    \[ M_1 = \dfrac{\text{moles}_1}{\text{liter}_1}\]

    rearrange the equations to find moles:

    \[ \text{moles}_1 = M_1 \text{liter}_1 \]

    and

    \[ \text{moles}_2 = M_2 \text{liter}_2 \]

    What stayed the same and what changed between the two solutions? By adding more water, we changed the volume of the solution. Doing so also changed it's concentration. However, the number of moles of solute did not change. So,

    \[moles_1 = moles_2\]

    Therefore

    M1 × litre1 = M2 × litre2

    Which we will usually express as:
    \[M_1V_1= M_2V_2 \]

    where

    • \(M_1\) and \(M_2\) are the concentrations of the original and diluted solutions and
    • \(V_1\) and \(V_2\) are the volumes of the two solutions

    Preparing dilutions is a common activity in the chemistry lab and elsewhere. Once you understand the above relationship, the calculations are easy to do.

    Example 1

    What volume of concentrated sulfuric acid, 18.0 M, is required to prepare 5.00 L of 0.150 M solution by dilution with water?

    Solution

    In a dilution question there are 4 variables: \(M_1\), \(V_1\), \(M_2\) and \(V_2\). You will know three of these values and have to calculate the fourth. Organizing your information is the key to correctly answering dilution questions. The most common mistake students make is in incorrectly pairing up concentrations and volumes. Take time to make sure you do not make this mistake.

    • \(M_1 = 18.0\; M\)
    • \(M_2= 0.150\; M \)
    • \(V_1 = ? \)
    • \(V_2 = 5.00\; L\)

    Set up the formula, and rearrange to solve for the unknown, V1:

    \[M_1V_1= M_2V_2 \nonumber \]

    \[(18.0)(V_1)= (0.150)(5.00) \nonumber \]

    \[ V_1 = \dfrac{ (0.150)(5.00) }{18.0} \nonumber\]

    \[ V_1 = 0.0417 \; L \nonumber = 41.7\; mL\]

    Stock Solutions

    It is often necessary to have a solution whose concentration is very precisely know. Solutions containing a precise mass of solute in a precise volume of solution are called stock (or standard) solutions. To prepare a standard solution a piece of lab equipment called a volumetric flask should be used. These flasks range in size from 10 mL to 2000 mL are are carefully calibrated to a single volume. On the narrow stem is a calibration mark. The precise mass of solute is dissolved in a bit of the solvent and this is added to the flask. Then enough solvent is added to the flask until the level reaches the calibration mark.

    Example 2: Preparation of Solution from a Stock Solution

    Describe how you would prepare 500.0 mL of a 0.100 M standard solution of KNO3

    Solution

    We must first determine the mass of KNO3 that will be needed to prepare 500.0 mL (0.500 L) of a 0.100 M solution. The molar mass of KNO3 is 101.1 g·mol-1:

    \[ g= \left( \dfrac{101.1 \; g}{ \cancel{mol}} \right) \left( \dfrac{0.100 \cancel{mol}}{\cancel{L}}\right) (0.500 \cancel{L}) = 5.05\;g\]

    To prepare the standard solution we will carefully measure out 5.05 g of KNO3 and dissolve it in some water (perhaps 200 mL of water). We pour this solution into a 500 mL volumetric flask and add just enough water reach the calibration mark on the flask.


    2.4 Dilutions of Standard Solutions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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