4. Data in the following table were obtained for the titration of a 0.297-g samp

Question

4. Data in the following table were obtained for the titration of a 0.297-g sample of a solid, monoprotic weak acid with a 0.150 M NaOH solution. Plot (at right opH (ordinate) vs. VNaoH (abscissa) a. To determine the molar mass and the pKa of a solid, monoprotic weak acid, a titration of the weak acid with a standardized NaOH solution provided the following data in the table. VNso added (mL) pH 1.96 2.22 12 2.00 4.00 2.46 7.00 10.00 3.06 12.00 3.29 14.00 3.60 16.00 4.26 17.00 11.08 18.00 11.67 20.00 12.05 25.00 12.40 2 S. pt 7.7 lo, SAL Experiment 18 233 mdan Muss OI 247

Explanation / Answer

LEt's do this by the numbers.

#4: the final volume is the volume at the end of reaction, which is 25 mL (according to the graph).

#5: This volume is the volume at equivalence point, which is in 16.5 mL (according to your graph)

#7: moles = 0.150 * 0.0165 = 0.002475 or 0.00248 moles

This moles are the same moles of acid.

#9: molar mass, you did that already in the sheet, however a little fixing:

MM = 0.297 / 0.002475 = 120 g/mol

11. This volume comes from the graph and it's in 10 mL (as you have mentioned in your graph)

#12. The pKa is the same as pH in this point, so pH = pKa = 3.06

If you need something else or something to be fixed, tell me in a comment and I'll do it.

Hope this helps

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