Volumetric Analysis: Acid-Base (chpt. 13)

Calculations: 1. Concentrations of solutions 2. Dilution of solutions 3. Reaction between solution and solid (finding mass) 4. Volumetric problems (finding mol/L, g/L, % w/v) 5. Water of crystallisation

Solution

Completely perfect mixture of a solute and solvent, where the particles of the solute are uniformly distributed throughout the solvent, ie mixture is homogeneous

Conversions

• g/L to mol/L

• no. particles to mol/L

• g/L to mol/L = divide by molecular mass (multiply for opp)

• no. particles to mol/L = divide by avagadro’s (multiply for opp)

Concentrated

Large amount of solute relative to the quantity of solvent

Dilute

Small amount of solute relative to the quantity of solvent

Concentration of a solution

Amount of solute that is dissolved in a given volume of solution

Expressions of percentage of solute

• Weight per weight (w/w) ie g per g

• Weight per volume (w/v) ie g per cm³ [g/100cm³]

• Volume per volume (v/v) ie cm³ per cm³

Find mass for volume when percentage of solute given

0.85% w/v salt. Find mass of sodium chloride needed to make 250cm³

1. Find mass per 100cm³

   0.85% = 0.85g NaCl per 100cm³ solution

2. Convert to mass per 250cm³

   250/100 = 2.5, therefore 0.85 × 2.5 = 2.125g NaCl

Find percentage for mass when mass of solute given

65g solution contains 8g solute, express concentration of solution as % w/w

Find percentage by dividing solute by solution and multiply by 100

8/65 × 100 = 12.31% w/w

Find concentration in ppm (parts per million)

0.005g calcium carbonate per 100cm³ water, find concentration of calcium carbonate in ppm

1. Find per litre (as 1ppm = 1mg/L)

   0.005g x 10 = 0.05g/L

2. Convert g/L into mg/L by multiplying by 1,000 (as 1,000 × 1,000 = 1,000,000 allowing ppm)

   0.05 × 1,000 = 50mg/L = 50ppm

Molarity of a solution

Number of moles of solute per litre of solution

One molar (1M) solution

Solution that contains one mole of the solute dissolved in one litre of solution

Decimolar solution

Solution that contains one tenth of a mole per litre (0.1M)

To find no. moles

n=MV (moles = molarity x Volume in L)

Concentration (formula)

Molarity x 1,000/volume

Find no. moles given volume and molarity

Moles HCl present in 30cm³ 0.2M HCl

Use equation, moles = volume x molarity/1,000

30 × 0.2/1,000 = 0.006M

How to calculate mass of a solid in a reaction

Mass of magnesium that reacts with 50cm³ H₂SO₄

Mg + H₂SO₄ → MgSO₄ + H₂

1. Create equation if not already given

2. Find relative atomic masses, then finding g/cm³ of compound trying to find by dividing by given mass’ re. atomic mass and multiplying by g needed

   50 × 0.5/1,000 = 0.025M

3. Look at balanced equation to see how many moles of each substance react, then multiply by by relative molecular mass of solid trying to find to find mass

   1M sulfuric acid reacts w 1M magnesium, therefore 1:1

   0.025 × 24 = 0.6g magnesium

Concentration rule when working with dilution

Moles of solute in diluted solution = moles of solute in concentrated solution

Find volume of one solution to make up a different concentration of another solution

Find volume of 12M HCl solution required for 500cm³ of 3M HCl

Use concentration rule

Vol(diluted) x M(diluted) = Vol(concentrated) x M(concentrated)

500 × 3 = Vol(concentrated) x 12, Vol = 500 × 3/12 = 125cm³

To find mol/cm³

Moles/cm³ used

To prepare standard solution sodium carbonate

Steps

1. Known mass anhydrous sodium carbonate weighed on clock glass, dissolved in deionised water adding washings, and added to volumetric flask

2. Filled with deionised water until bottom of meniscus filled at eye level, invert about 20 times to mix thoroughly

Notes

1. This is a known concentration, ie standard solution, as moles per litre of solution can be calculated

2. Must be deionised water, tap water contains many dissolved substances that could interfere w solution and affect results

Titration

Laboratory procedure where a measured volume of one solution is added to a known volume of another solution until the reaction is complete

Equivalence point

Stage when two solutions in a titration just react completely with each other

End point

When equivalence point is shown by an indicator, and is therefore the end of a titration

To use standard solution of sodium carbonate solution to standardise a given hydrochloric acid solution

Steps

1. Pipette, burette and conical flask are washed with deionised water, then solution that the pipette and burette are to contain

2. 25cm³ sodium carbonate pipetted from beaker into conical flask (beaker should be dry before decanted sodium carbonate from volumetric flask). A few drops methyl orange indicator added to conical flask

3. Burette clamped vertically to retort stand filled to above zero mark with dilute hydrochloric acid solution. Funnel removed, then tap filled by releasing acid until at the zero mark at the bottom of the meniscus.

4. White tile is added under conical flask. Rough titre begins where acid let through while swirling flask until methyl orange becomes a pink colour. Two further more accurate titrations are completed which should agree to within 0.1cm³

Explanations

• Only small amount of indicator should be added as acid-base indicators are usually weak acids/bases, which with too much could lead to inaccurate result

• Methyl orange is suitable as titrating a strong acid against a weak base

• Acids in the attic, base in the basement, as bases could cause the tap of the burette to stick, with sodium hydroxide attacking glass and absorbas carbon dioxide forming a crust of sodium carbonate around the tap.

• Washing down the side of flask ensures no drops of acid stuck to flask, with deionised water which doesn’t affect no. moles in sodium carbonate

• Not washing with solution it is to cotain can have dilution occurring after required amount of solution is added, leading to inaccurate results

• Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Calculating unknown concentration of a solution from titration data - Formula Method

1.45g sodium carbonate made to solution of 250cm³, 25cm³ titrated against 19.55cm³ hydrochloric acid

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

1. If not given, find molarity of given solution

   0.055M Na₂CO₃

2. Use concentration of one to find concentration of other

   2:1 ratio. using equation ↓

   V₁ = 19.55, V₂ = 25, M₁ = ?, M₂ = 0.055, n₁ = 2, n₂ = 1

   19.55 x M₁/2 = 25 × 0.055/1 → M₁ = 0.141mol/L, 0.141 × 2(1.45) = 5.147g/L

To make up an approx. 0.1M sodium hydroxide and standardise with standard hydrochloric acid solution, hence preparing sample sodium chloride (Washing Soda)

Steps

1. Sodium hydroxide (1g) dissolved in deionised water and solution made up to 250cm³, then invert about 20 times to dissolve

2. Pipette, conical flask and burette rinsed with deionised water and then (bar flask) the solution it is to contain

3. 25cm³ sodium hydroxide solution pipetted into conical flask and a few drops methyl orange added

4. Burette filled with dilute hydrochloric solution and white tile place under conical flask. 3 titres, 2 agreeing within 0.1cm³

5. Sample sodium chloride is then obtained by repeating titration without indicator by adding same amount of hydrochloric acide solution. This is then left to evaporate on evaporating dish and hotplate

Explanations & Observations

• Methyl orange goes orange to pink as solution in flask becomes acidic

• White crystals are formed on evaporating dish, which can be left to dry, showing crystals of sodium chloride formed

• Sodium hydroxide can cause severe burns, ppe must be worn

• Only approximate of moles of sodium hydroxide can be made up as it is deliquescent (absorbs so much water vapour from air that it becomes a solution)

To standardise

To obtain the exact concentration of a substance in a solution

To determine concentration ethanoic acid (acetic acid) in vinegar

Steps

1. Two 25cm³ samples vinegar added to 250cm³ volumetric flask and filled with deionised water (diluted by factor of 5)

2. Pipette, burette, and conical flask washed w deionised water and rinsed (bar flask) w solution to contain

3. 25cm³ standard sodium hydroxide solution pipetted into conical flask and 3 drops phenolphthalein indicator added. Burette filled with diluted vinegar, white tile placed. 3 titres, 2 within 0.1cm³

Explanations & Observations

• Phenolphthalein is pink as is pink in basic solution, becoming colourless at end point as is colourless when acidic

• It is necessary to dilute vinegar as concentration sodium hydroxide solution used is small also so as to avoid having to use large volume of sodium hydroxide solution and a visible amount of vinegar, which would contribute to a higher percentage error

Titration Formula

volume 1 (V₁) x molarity 1 (M₁)/no. moles 1 (n₁) = volume 2 (V₂) x molarity 2 (M₂)/no. moles 2 (n₂)

To determine amount of water of crystallisation in hydrated sodium carbonate (washing soda)

Steps

1. 5g washing soda weighed on clock glass, then dissolved in deionised water and transferred to 250cm³ volumetric flask with washings, then filled and inverted about 20 times

2. Pipette, burette, and conical flask washed w deionised water then (bar flask) solution to contain

3. 25cm³ sodium carbonate solution pipetted into conical flask from beaker and 3 drops methyl orange indicator added. Burette filled with hydrochloric acid solution. 3 titres