Question

The data for the titration of 25.00 mL of Unknown Acid Sample #80 with 0.1508 M NaOH (aq) was plotted on the graph below.

To help with your analysis we have summarized some of the key numerical values that were determined from the original data - using the methods on Page 13 of your lab manual to graphically determine the equivalence point.

	Unknown #27
Volume of NaOH added at the potentiometric endpoint (mL)	21.25 mL
Volume of NaOH added at the half-equivalence point (mL)	10.63 mL
pH at the half-equivalence point	4.90
pH at the start of the titration	3.23

Report your calculated original (stock) acid concentration (in mol/L) below, using the appropriate number of significant figures. Do not use scientific notation!

Your Answer:

Answer 1

Let HA be the unknown acid. HA + NaOH → Na A+ H₂O Molarity Eqh is : Mive = Mave - CHA) (NaOH) where, Mi= Molar concentration of HA = ? V = Volume of Acid = 25.00 mL . M2 = Molarity of NaOH = 0.1508 molle V₂ = Volume of NaOH at Equivalence point = 21.25 ml o Molarity of Unknown acid = Mi= M2 Ve = 0.1508 mol/L X 21. 25 ml 95.00 mL - = 0.1282 mol/L Mi