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Essay text:
Average titre = 24.80 cm3.
Moles of acid used = 5x10-3
Acid neutralised by 24.80 cm3 NaOH
Therefore concentration of NaOH = (5x10-3 x 1000)/ 24.8
=0.2016129…
Diluted by factor of 10, so: 0.2016129…x10 = 2.01 mol dm-3.
Method 2- ?H of neutralisation
• Transfer 25cm3 of NaOH into a polystyrene cup with a lid, and put the cup into a plastic beaker.
• Take the temperature of the contents of the cup using a thermometer accurate to 0.1 cm3.
• Add excess HCl and replace the lid of the cup.
• Stir the contents constantly until the maximum temperature has been reached.
• Record the temperature change...
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