Dilution – HSC Chemistry

 

This is part of Year 11 HSC Chemistry course under the topic of Molarity.

HSC Chemistry Syllabus

  • Manipulate variables and solve problems to calculate concentration, mass or volume using:
– Dilutions (number of moles before dilution = number of moles in sample after dilution). 
    • Conduct an investigation to make a standard solution and perform a dilution. 

    Dilution Theory & Calculation Examples

    What is Dilution?

    Dilution refers to the process of decreasing the concentration of a solution by adding more volume of solvent to the existing solution. In most cases, this solvent is water.

    The most important principle to understand in dilution of a solution is that the number of moles of solute remains unchanged with the addition of solvent.

     

    Dilution in a beaker

     

    Therefore, the number of moles of solute before dilution equals the number of moles of solute after dilution, as given by:

     

    $$n_{\text{before}} = n_{\text{after}}$$

     

    Since the number of moles of a substance is given by:

     

    $$n = c \times V$$

     

    We get:

       

    `c_1 \times V_1 = c_2 \times V_2`

     

    where

    • `c_1` is the concentration of the original solution before dilution in mol L–1
    • `V_1` is the volume of the original solution before dilution in litres (L)
    • `c_2` is the concentration of the new diluted solution  in mol L–1
    • `V_2` is the new volume of the diluted solution in litres (L)

    Example 1

    Water is added to 100.0 mL of 0.250 M sodium chloride solution until the volume reaches 1.50 L. Calculate the new concentration of this solution.
     

      Solution:

      The number of moles of sodium chloride before dilution is given by:

      $$n = 0.250 \times 0.100 = 0.0250 \text{ mol}$$

       

      After dilution, the number of moles of NaCl remains the same.

      Therefore,

      $$c_{\text{new}} = \frac{0.0250}{1.50} = 0.0167 \text{ mol/L (3 s.f.)}$$

       

      Alternatively, we can use the dilution equation:

      Since `C_1` = 0.25 M, `V_1` = 0.1 L, and `V_2` = 1.5 L, we can rearrange the equation to solve for `C_2`:

       

      $$c_2 = \frac{c_1 \times V_1}{V_2} = \frac{0.25\, \text{mol/L} \times 0.1\, \text{L}}{1.5\, \text{L}} = 0.0167\, \text{mol/L}$$

      Example 2

      How many litres of water needs to be added to a 500.0 mL solution of 0.10 M hydrochloric acid solution to make a 0.010 M solution?

       

        Solution:
        The number of moles of HCl is given by:
         
        $$n = c \times V = (0.10)(0.5)$$
        $$n = 0.050 \text{ mol}$$
         
        The new final volume of solution is given by:
         
        $$V = \frac{n}{c}$$
        $$V = \frac{0.050}{0.010}$$
        $$V = 5.0 \text{ L}$$
         
        $$\text{Volume added} = 5.0 - 0.5  = 4.5 \text{ L (2 s.f.)}$$

        Example 3 

        200.0 mL of water is added to 30.0 mL of a copper sulfate solution. The final concentration of the solution is 0.120 mol L–1.

        What is the original concentration of the copper sulfate solution?

         

          Solution:
          The number of moles of copper sulfate is given by:
           
          $$n = c \times V = (0.120)(0.200 + 0.030)$$
          $$n = 0.0276 \text{ mol}$$
          The original solution would have contained the same number of moles of copper sulfate.
           

          $$c = \frac{n}{V} = \frac{0.0276}{0.030}$$

          $$c = 0.920 \text{ mol/L (3 s.f.)}$$

          Example 4

              2.00 g of solid NaOH is added to 300.0 mL of water. 150.0 mL of this solution was transferred to a new beaker and diluted with 250.0 mL of water.

              What is the final molarity of the NaOH solution?

               

              Solution:

              Let's first find the number of moles of NaOH added:

               

              $$n = \frac{m}{MM} = \frac{2.00}{22.99 + 16.00 + 1.008}$$

              $$n = 0.0500 \text{ mol}$$

               

              Initial molarity:

               

              $$c = \frac{n}{V} = \frac{0.0500}{0.300}$$

              $$c = 0.167 \text{ mol/L}$$

               

              The number of moles in the 150.0 mL of solution that is transferred to a new beaker is not the same as the original solution. However, this portion of the solution has the same concentration.

               

              $$n_{\text{transferred}} = c \times V = 0.167 \times 0.150$$

              $$n_{\text{transferred}} = 0.025 \text{ mol}$$

               

              Lastly, during dilution with 250.0 mL of water, the number of moles of NaOH remains unchanged. The final volume = 150 + 250 = 400.0 mL = 0.400 L.

               

              $$c = \frac{n}{V}$$

              $$c = \frac{0.025}{0.400}$$

              $$c = 0.0625 \text{ mol/L (3 s.f.)}$$

               

               

              RETURN TO MODULE 2: INTRODUCTION TO QUANTITATIVE CHEMISTRY