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

4. Find the molarity of the vinegar for each of your three vinegar titrations, then find the average molarity of the acid.

5. Find the percent deviations for your three vinegar molarities for the vinegar titrations.

6. From the moles of acetic acid for each vinegar titration, find the mass of acetic acid for that titration

Answer 1

Vinegar:	Titration 1	Titration 2	Tirtration 3
1. Initial V of Acid	.01 mL	5.56 mL	11.32 mL
2. V of Acid	5.56 mL	10.71 mL	16.02 mL
3.Inital V of Base	.01 mL	27.12 mL	3.01 mL
4.Final V of Base	22.23 mL	44.71 mL	24.90 mL
V of Acid taken for titration, mL (Line 2 –line 1)	5.55	5.15	4.70
V of NaOH used for titration, mL (line 4 – line 3)	22.22	17.59	21.89
Ratio VNaOH / Vvinegar	4.054	3.41	4.65

C_vinegar = C_NaOH x V_NaOH / V_vinegar = C_NaOH x Ratio V_NaOH / V_vinegar

(see Table for Ratio V_NaOH / V_vinegar values)

I don’t find here the value of C_NaOH . I don’t understand what this means:

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

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

4. answer with the average value of the 3 results

5. Calculate a standard deviation s of the 3 results.

Percent deviation is 100 s / mean value