To standardize your NaOH solution, you titrate 20.0 mL 0.0769 M HCI solution, me
ID: 493446 • Letter: T
Question
To standardize your NaOH solution, you titrate 20.0 mL 0.0769 M HCI solution, measuring the pH after each small addition of NaOH. Below is the graph you obtain experimentally. The equivalence point occurs where there is a steep rise in pH. What molar quantity of HCl is being titrated? What molar quantity of NaOH has been added at the equivalence point? What volume of NaOH solution has been added at the equivalence point? Expanded view of the equivalence point: What is the concentration of the NaOH solution?Explanation / Answer
GIven
Molarity of HCl = 0.0769 M(mol/L)
Volume of HCl = 20 ml = 0.02 L
No of moles of HCl = Volume * Molarity = 0.0769 mol/L * 0.02 L = 1.54 * 10-3 moles
Molar quantity of HCl being titrated = 1.54 * 10-3 moles Answer
At equivalence point Molar quantity of HCl being titrated will be equal to Molar quantity of NaOH used up
Molar quantity of NaOH added at equivalence point = 1.54 * 10-3 moles Answer
Volume of NaOH at equivalence point = 23.9 ml = 0.0239 L Answer
Molarity of NaOH used = Molar quantity of NaOH at equivalence point / Volume of NaOH used at equivalence point
= 1.54 * 10-3 moles / 0.0239 L = 0.0643 M
Molarity of NaOH used = 0.0643 M Answer
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