What will be the volume of concentrated H2SO4 that must be added to 155 mL of 0.1996 N H2SO4 in a 500 mL volumetric flask so that the resulting solution will be 6.5210 M?
pls. explain.
To find the volume of concentrated H2SO4 required, we need to use the formula for dilution:
M1V1 = M2V2
where:
M1 = initial concentration of the solution to be diluted
V1 = initial volume of the solution to be diluted
M2 = final desired concentration of the diluted solution
V2 = final volume of the diluted solution (which is the sum of the initial volume and the volume of the concentrated H2SO4 added)
Given data:
M1 = 0.1996 N
V1 = 155 mL
M2 = 6.5210 M
V2 = 500 mL
First, we need to convert the N (normality) unit to M (molarity) by dividing by the equivalent factor. For H2SO4, the equivalent factor is 2 because it can donate 2 moles of H+ ions per mole of H2SO4.
M1 = 0.1996 N / 2 = 0.0998 M
Now we can plug the values into the formula and solve for V2:
(0.0998 M)(155 mL) = (6.5210 M)(500 mL + V2)
Rearranging the equation to solve for V2, we get:
V2 = [(0.0998 M)(155 mL)] / (6.5210 M) - 500 mL
Calculating this equation will give us the volume of concentrated H2SO4 required to achieve the desired concentration.
Let's compute:
V2 = [(0.0998 M)(155 mL)] / (6.5210 M) - 500 mL
V2 = 2.378 mL
Therefore, approximately 2.378 mL of concentrated H2SO4 needs to be added to a 155 mL solution of 0.1996 N H2SO4 to achieve a final concentration of 6.5210 M in a 500 mL volumetric flask.