25 ml of milk were titrated with a 0.124M KOH solution of potassium hydroxide. If 4.5 ml of KOH solution were required to reach the titration endpoint, with a percentage (% w / v) lactic acid present in the milk? (MM = 90.08g / mol)
Lactic acid is a monoprotic acid which I will designate as HL.
HL + KOH ==> KL + H2O
Then mols KOH used = M x L = ?
mols HL = mols KOH since the equation is 1:1 in HL:KOH
grams HL = mols HL x molar mas HL = ?
Then % w/v = (grams HL/25 mL)*100 = ?
To find the percentage of lactic acid (% w/v) in the milk, we need to calculate the concentration of lactic acid and then convert it to a percentage.
Step 1: Calculate the number of moles of KOH used in the titration.
Given:
- Volume of KOH solution used = 4.5 ml
- Concentration of KOH solution = 0.124 M
Use the formula:
moles of solute = concentration × volume (in liters)
Converting the volume from milliliters to liters:
4.5 ml = 4.5/1000 = 0.0045 L
Number of moles of KOH = 0.124 M × 0.0045 L = 0.000558 moles
Step 2: Calculate the number of moles of lactic acid reacted.
From the balanced equation of the reaction between KOH and lactic acid, we know that the stoichiometric ratio is 1:1. This means that for every mole of KOH used, one mole of lactic acid reacts.
Since 0.000558 moles of KOH reacted, the same amount of moles of lactic acid must be present in the milk.
Number of moles of lactic acid = 0.000558 moles
Step 3: Calculate the mass of lactic acid present.
Given:
- Molar mass of lactic acid (MM) = 90.08 g/mol
Use the formula:
mass = number of moles × molar mass
Mass of lactic acid = 0.000558 moles × 90.08 g/mol = 0.0502 g
Step 4: Calculate the percentage (% w/v) of lactic acid in the milk.
Given:
- Volume of milk used = 25 ml
Use the formula:
% w/v = (mass of lactic acid / volume of milk) × 100
% w/v = (0.0502 g / 25 ml) × 100 = 0.2016 % w/v
Therefore, the percentage (% w/v) of lactic acid present in the milk is approximately 0.2016%.