The average NaOH molarity was .153. Click the2 Derivative button (top right) to

Question

The average NaOH molarity was .153.

Click the2 Derivative button (top right) to display your data as shown in you lab book 1t Derivative 2"d Derivative Draw Lines Mouse Coordinates: 17.856,11.338 Titration of Citric Acid Trial 2 Line 1 12 17.900,11.300 10 9 6 5 0 10 15 20 25 Volume (mL) of base at the Equivalence Point Try 1: 15.04 Try 1: 17.18 You MUST view EACH graph before submitting this answer 0.5 0.5 0.5 0.5 8. Using your average NaOH molarity, determine the molarity (M) of the Citric Acid solution NG 0.5 NG 0.5

Explanation / Answer

M1V1 = M2V2
M1 = Molarity of citric acid in mol/L ; V1 = Volume of the acid titrated in L
M2 = Molarity of NaOH = 0.153 mol/L ; V2 = 17.18 mL = 17.18/1000 = 0.01718L
Volume of the acid titrated is not given in the question however use the above formula
to calculate M1

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