Dilutions

I. Diluting Solutions

                A. When 2 solutions are mixed the concentration chenges

                B.  Dilution: process of decreasing the concentration of a solution by adding solvent (usually water)

                                a. The amount of solute doesn't change, only the volume changes

                                                i. v₁=v₂

                C. Because concentration is [mol/L] we can write:

                                a. C= n/v              or           n=C/v   so C₁v₁=C₂v₂

Ex. Determine the concentration when 100mL of 10.0M HCL is diluted to a final volume of 400mL

C₁= 0.1M

V₁=100mL

C₂=  ?

V₂= 400mL

C₁v₂=C₂v₂

C₁v₁ / v₂= C₂

(0.1M)(100mL) / (400mL) = 0.025 mol/ mL

Ex. How much water must be added to 10.0mL of 10.0M Na₂SO₄ to give a solution with a concentration of 0.50M?

C₁=10.0M

V₁= 10.0mL

C₂=0.50M

V₂=?

C₁V₁=C₂V₂

C₁V₁ / C₂= V₂

(10.0M)(10.0mL) / (0.5M) = 200mL

ΔV = V_f - V_i

ΔV = 200 - 10

ΔV = 190 mL of water needs to be added

Ex. How much water must be evaporated from 2.00L of 0.250M KCl solution for the final concentration to be 2.75M?

C₁=0.250M

V₁= 2.00L

C₂=2.75M

V₂=?

C₁V₁=C₂V₂

C₁V₁ / C₂= V₂

(0.250M)(2.00L) / (2.75M) = 0.182L

ΔV = V_f - V_i

ΔV = 0.182 - 2.00

ΔV = 0.182 - 200

ΔV= -1.82

Ex. [100 mL of 0.250M]_a sodium nitrate is mixed with [200mL of 0.100M]_b sodium nitrate. Find the concentration.

-Find the mol and volume of each solution separately

Solution a.  (0.100L) x (0.250mol / 1L) = 0.025 mol

Solution b. (0.200L) x (0.100mol / 1L) = 0.0200 mol

-Add together

mol_tot = 0.025mol + 0.0200mol = 0.0450mol

volume_tot= 100mL + 200mL = 300mL

(mol_tot) / (Volume_tot) = Molarity_tot

(0.0450mol) / (0.300L) = 0.150M

Ex. What mass of silver nitrate is needed to make 75.0mL of a 0.130M solution?

(0.130mol / 1L) x (0.075L) x (169.9g / 1mol) = 1.66g