Nigel Baldwin | Titration of sodium hydrogen sulfate with sodium hydroxide C0166 @nigelbaldwin3184 | Uploaded May 2020 | Updated October 2024, 2 hours ago.
High School Chemistry

A 50ml burette was filled with sodium hydroxide of unknown concentration.

The initial reading from the burette was recorded in a lab book.

25ml of 0.11M sodium hydrogen sulfate solution was placed in a 250ml conical flask using a volumetric pipette.

2 drops of phenolphthalein indicator was added to the sodium hydrogensulfate.

Sodium hydroxide was added to the conical flask from the burette until a faint pink colour was observed in the flask which persisted for at least 30 seconds.

The final reading was recorded in the lab book.

The titration volume of sodium hydroxide solution added was worked out; final reading - initial reading.

Two concordant titration volumes were observed, i.e. two titration volumes within 0.1ml of each other.

The concentration of the sodium hydroxide solution may be calculated.

Calculation for the concentration of sodium hydroxide using the second titration volume.

Titration volume = 23.35ml

Number of moles of sodium hydrogen sulfate, n = c x V

c = concentration = 0.11 mol / dm3
V = volume pipetted = 0.025 dm3

n = 0.11 x 0.025 = 0.00275 mol

The number of moles of sodium hydroxide = number moles sodium hydrogen sulfate

This is because they react in a 1 mol + 1 mol stoichiometric ratio in the reaction equation.

Therefore, the number of moles of sodium hydroxide = 0.00275 mol

Concentration of sodium hydroxide, c = n / V

c = 0.00275 / 0.02635

c = 0.104364326 mol / dm3

c = 0.10 mol / dm3 to 2 significant figures