The molarity of hydrochloric acid waste water sample is determined by titration. At the start of the tiration the volume of 0.015M NaOH in a buret is read. The recorded value is 0.02mL. The tiration proceeds as the NaOH solution is dispensed to a 50.0mL sample of waste water. The phenolphthalein indicator changed color when the buret read 32.32 mL. What is the molarity of the HCl in the wastewater sample?
Titration Reaction: HCl (aq)+NaOH (aq)-->NaCl+H2O(l)
Titration Reaction: HCl (aq)+NaOH (aq)-->NaCl+H2O(l)
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From the reaction equation you know that 1 mol NaoH reacts with 1 mol HCl.
Volume of NaOH used: 32.32 - 0.02 = 32.30mL NaOH used
Molarity of NaOH = 0.015M
Mol of NaOH in 32.30mL of 0.015M solution = 32.3/1000*0.015 = 0.0004845 mol NaOH
Therefore you know that 50.0mL of water contained 0.0004845 mol HCl
1000mL of waste water will contain : 1000/50.0 * 0.0004845 = 0.00969 mol HCl per litre of water.
Molarity of HCl in waste water = 0.00969M
Volume of NaOH used: 32.32 - 0.02 = 32.30mL NaOH used
Molarity of NaOH = 0.015M
Mol of NaOH in 32.30mL of 0.015M solution = 32.3/1000*0.015 = 0.0004845 mol NaOH
Therefore you know that 50.0mL of water contained 0.0004845 mol HCl
1000mL of waste water will contain : 1000/50.0 * 0.0004845 = 0.00969 mol HCl per litre of water.
Molarity of HCl in waste water = 0.00969M
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50 ml of the waste water sample were neutralised by 32.3 ml 0.015M NaOH
moles naOH used = 32.3 x 0.015/1000 = 4.845 x 10^-4
Since 1 mole HCl reacts with 1 mole NaOH then 50 ml waste water sample contains 4.845 x 10^-4 moles HCl or 4.845 x 10^-4 x 1000/50 = 0.00969M
moles naOH used = 32.3 x 0.015/1000 = 4.845 x 10^-4
Since 1 mole HCl reacts with 1 mole NaOH then 50 ml waste water sample contains 4.845 x 10^-4 moles HCl or 4.845 x 10^-4 x 1000/50 = 0.00969M