Question

Molarity of standard NaOH solution: x/1 H 2/3 Unknown + 2/3 NaOH -> 2/3 H2O +Na 2/3 Acetate

Vinegar:	Titration 1	Titration 2	Tirtration 3
Initial V of Acid	.01 mL	5.56 mL	11.32 mL
Final V of Acid	5.56 mL	10.71 mL	16.02 mL
Inital V of Base	.01 mL	27.12 mL	3.01 mL
Final V of Base	22.23 mL	44.71 mL	24.90 mL

Unknown Acid code and number of reactive hydrogens: H+3

Unknown Acid	Titration 1	Titration 2 (Error)	Titration 3	Titration 4
Mass of Acid	.1024 g	.1028g	.1021g	.1023g
Initial V of Base	24.90 mL	37.80 mL	45.5 mL	9.24 mL
Final V of Base	37.80 mL	45.5 mL	9.24 mL	17.80 mL

10. Find the moles of base (NaOH) for each of your unknown acid titrations.

11. Find the balanced equation using the number of reactive hydrogens (call an acid with one reactive hydrogen HA, one with two reactive hydrogens H₂A, and so on).

12. Find the moles of unknown acid for each of your three unknown acid solutions.

13. Use the mass of unknown acid and the moles of unknown acid to find the formula mass of your unknown acid for each of your three unknown acid titrations.

Answer 1

Unknown Acid	Titration 1	Titration 2 (Error)	Titration 3	Titration 4
1. Mass of Acid	.1024 g	.1028g	.1021g	.1023g
2. Initial V of Base	24.90 mL	37.80 mL	45.5 mL	9.24 mL
3.Final V of Base	37.80 mL	45.5 mL	59.24 mL	17.80 mL
4 . VNaOH used in titration, mL (line 3 – lin2)	12.90	7.70 error	13.74	8.56 error
5.   n, mol of NaOH used	……	….	….	…..
6. moles of unknown acid Line 5 divided by 3	…..	……	…	…..
7. molar mass of H3A (line 1 / line 6)	….	Not used	……	Not used

10.

For line 5:

n, mol of NaOH used = V_NaOH (from line 4) x C_NaOH

(I don’t have here the value of NaOH)

11.

H₃A + 3 NaOH = Na₃A + 3 H₂O

12.

“the moles of unknown acid for each of your three unknown acid solutions” are calculated as

Line 5 divided by 3

(the molar ratio H₃A / NaOH   is 1:3)

13.

If titration 2 is an error. Titration 4 is also an error.

Use only titrations 1 and 3 to calculate the molar mass. Calculate the average value.