Question

Data: Trial 1 Trial 2 Trial 3 Initial Volume NaOH 10.00mL 10.00mL 10.00mL Final Volume NaOH 1.70mL 2.10mL 1.90mL Total NaOH Volume 8.30mL 7.90mL 8.20mL Volume of Unknown Acid 5.00mL 5.00mL 5.00mL Analysis (Calculations): 1. Determine the average volume of NaOH used in each trial (add the three volumes, divide by exactly 3—watch your significant figures!): 8.30mL + 7.90mL + 8.20mL = 24.4/3 = 8.13 mL 2. You do not know which acid you were given, but it was mono-protic. Therefore, your mole ratio of NaOH to Acid is 1:1. There was no question here. 3. Complete your calculations (show the dimensional analysis in your report).: 1. The moles of sodium hydroxide consumed (using the average volume). 2. The moles of monoprotic acid in your original solution. 3. The molarity of your unknown acid (this is your "final answer").

          Data:

Trial 1 Trial 2 Trial 3
Initial Volume NaOH 10.00mL 10.00mL 10.00mL
Final Volume NaOH 1.70mL 2.10mL 1.90mL
Total NaOH Volume 8.30mL 7.90mL 8.20mL
Volume of Unknown Acid 5.00mL 5.00mL 5.00mL

Analysis (Calculations):

1. Determine the average volume of NaOH used in each trial (add the three volumes, divide by exactly 3—watch your significant figures!):
8.30mL + 7.90mL + 8.20mL = 24.4/3 = 8.13 mL

2. You do not know which acid you were given, but it was mono-protic. Therefore, your mole ratio of NaOH to Acid is 1:1.

There was no question here.

3. Complete your calculations (show the dimensional analysis in your report).:
1. The moles of sodium hydroxide consumed (using the average volume).

2. The moles of monoprotic acid in your original solution.

3. The molarity of your unknown acid (this is your "final answer").

Data:

Trial 1 Trial 2 Trial 3
Initial Volume NaOH 10.00mL 10.00mL 10.00mL
Final Volume NaOH 1.70mL 2.10mL 1.90mL
Total NaOH Volume 8.30mL 7.90mL 8.20mL
Volume of Unknown Acid 5.00mL 5.00mL 5.00mL

Analysis (Calculations):

1. Determine the average volume of NaOH used in each trial (add the three volumes, divide by exactly 3—watch your significant figures!):
8.30mL + 7.90mL + 8.20mL = 24.4/3 = 8.13 mL

2. You do not know which acid you were given, but it was mono-protic. Therefore, your mole ratio of NaOH to Acid is 1:1.

There was no question here.

3. Complete your calculations (show the dimensional analysis in your report).:
1. The moles of sodium hydroxide consumed (using the average volume).

2. The moles of monoprotic acid in your original solution.

3. The molarity of your unknown acid (this is your "final answer").

Added by Cody S.

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