for full credit. Copy your answers to problems 1 through 3 info your lab noteboo

Question

for full credit. Copy your answers to problems 1 through 3 info your lab notebook for use when writing your lab report. To standardize your NaOH solution, you titrate 20.0 mL of 0.0769 M HC1 solution, measuring the pH after each small addition of NaOH. Below is the graph you obtain experimentally. The equivalence point occurs w here there is a steep rise in pH. What molar quantity of HC1 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? What is the concentration of the NaOH solution? Expanded view' of the equivalence point:

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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