Solution Concentration

Solution and Concentration

How can we express the amount of solute to solvent in How can we express the amount of solute to solvent in a solution? a solution?

• This is called the This is called the concentrationconcentration •a measure of the number of particles of the solute in the solvent

•Concentration =

Dilute Solutions

Small quantity of solute per unit volume of

solution.

Concentrated Solutions

Large quantity of solute per unit volume of

solution.

Ways to express concentration

3 ways of expressing concentration3 ways of expressing concentrationPercentages PPM or PPB concentrations molar concentrations and mass concentrations

Percentage Concentration

•% (w/w) = • W is weight same as mass

•% (w/v) =

•% (v/v) =

100)(

)(x

gsolutionmass

gsolutemass

100)(

)(x

mLsolutionvolume

gsolutemass

100)(

)(x

mLsolutionvolume

mLsolutevolume

Everyday Examples

• Vinegar is 5% v/v acetic acid– 5 mL of CH3COOH in 100 mL vinegar

• Hydrogen peroxide is 3% m/v– 3 g of H2O2 per 100 mL of solution

• Yogurt is 2% m/m– 2 g of milk fat per 100 g yogurt

Examples1. % V/V = 4.1 L / 55 L = 7.5% V/V

2. % W/V = 16 g / 50 mL = 32% W/V

3. % W/W = 1.7 g / 35.0 g = 4.9% W/W

More practice1. What is the % W/W of copper in an alloy when 10 g of

Cu is mixed with 250 g of Zn?

2. What is approximate % V/V if 30 mL of pure ethanol is added to 250 mL of water?

3. What is the % W/W if 8.0 g copper is added to enough zinc to produce 100 g of an alloy

% Concentration: % Mass Example

3.5 g of CoCl2 is dissolved in 100mL solution. Assuming the density of the solution is 1.0 g/mL, what is concentration of the solution in % mass?

%m = 3.5 g CoCl2

100g H2O

= 3.5% (m/m)

Low concentrations• For very dilute solutions, weight/weight

(w/w) and weight/volume (w/v)concentrations are sometimes expressed in parts per million (ppm).– Example:

• Toxic substances found in the environment• Chlorine in a swimming pool• Metals in drinking water

• Parts per million also can be expressed – as milligrams per liter (mg/L) (w/v)– As milligram per kilogram (mg/Kg) (w/w)

PPM learning check1. Question: A solution has a concentration of

1.25g/L. What is its concentration in ppm?• Convert the mass in grams to a mass in milligrams:

1.25g = 1.25 x 1000mg = 1250mg• Re-write the concentration in mg/L = 1250mg/L = 1250ppm

2. Question: 150mL of an aqueous sodium chloride solution contains 0.0045g NaCl. Calculate the concentration of NaCl in parts per million (ppm).

ppm = mass solute (mg) ÷ volume solution (L)mass NaCl = 0.0045g = 0.0045 x 1000mg = 4.5mg volume solution = 150mL = 150 ÷ 1000 = 0.150Lconcentration of NaCl = 4.5mg ÷ 0.150L = 30mg/L = 30ppm

Molar concentration• is able to compare the amount of solute (moles)

dissolved in a certain volume of solution.

• Molar concentration (mol/L) or also called Molarity (M)

= Moles of solute (n)/Volume of solution (V MUST be in litres)

• is the number of moles of solute in one litre of a solution. We use "M" to denote molar concentration and it has the units of"mol/L".

• C = n / V

• n = C x V

Concentration: Molarity ExampleIf 0.435 g of KMnO4 is dissolved in enough water to give 250. mL of solution, what is the molarity of KMnO4?

Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters — to give the molarity. Because 250. mL is equivalent to 0.250 L .

As is almost always the case, the first step is to

convert the mass of material to moles.

MM = 158.0 g/mol(39+55+(4X16)n= m/MM0.435 g KMnO4 / 158.0 g/ mol = 0.00275 mol KMnO4

Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 Mol/L = M

0.250 L solution

Molarity learning check1. How many moles of H2SO4 are there in 250mL of a 0.8M sulphuric acid

solution?2. If 20g of NaOH is dissolved in sufficient water to produce 500 mL of

solution, calculate the molar concentration in molarity.Questions1. What is the molarity of the solution formed by dissolving 80 g of

sodium hydroxide (NaOH) in 500 mLs of water? (Ar's: Na=23, O=16, H=1) Ans = 4 M

2. What is the molarity of the solution formed by dissolving 9.8 g of sulphuric acid (H2SO4) in 1000 cm3 of water? (Ar's: H=1, S=32, O=16) Ans = 0.1 M

3. What mass (g) of hydrogen chloride (HCl) is needed to make up 500cm3 of a solution of concentration 0.2mol/L? (Ar's: H=1, Cl=35.5) Ans = 3.65g

Preparing a Solution by Dilution

• Dilution = the process of decreasing the concentration of a solution, usually by adding more solvent

• Stock Solution = a solution that is in stock or on the shelf (i.e., available); usually a concentrated solution

Calculating the New Concentration of the Diluted Solution

C1 x V1 = C2 x V2

• C1 = initial concentration• V1 = initial volume (volume of stock solution

needed to make the concentration C2)

• C2 = final concentration

• V2 = final volume

Sample Problem• Water is added to 0.200L of 2.40mol/L NH3(aq)

cleaning solution, until the final volume is 1.000L. Find the molar concentration of the final, diluted solution.