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i am lost woth Titration, please help... i was able tondo my virtual lab and get the data needed... but i dont understand what to do on the remaining problems! please help!

Experiment 11 part 2 ACID BASE TITRATIONS OBJECTIVES 1. To learn general titration techniques 2. To generate a standardized t
Name Experiment 11 part 2 - report page (1 of 2) 1. Calculate the molarity of the unknown acid for trials 1-3 using dimension

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Answer #1

*CORRECTION: The volume of unknown acid in the table should be 25 mL. The 0.4 mL of the indicator should not be counted as an unknown amount of acid.  

Q1)

Trial 1

The volume of NaOH required for complete titration (V1) = 10.7 mL

Molarity of NaOH (M1) = 0.1 M

Volume of unknown acid(V2) = 25 mL

Molarity of unknown acid (M2) = to be found = x

so, application of dimensional analysis yields the formula

MV1 -factor x V2 n {Where n-factor of diprotic acid is 2}

Plugging the values

0.1M X 10.7mL C 2 x 25ml

therefore x = unknown concentration = 0.0214 M

Trial 2

The volume of NaOH required for complete titration (V1) = 10.7 mL

Molarity of NaOH (M1) = 0.1 M

Volume of unknown acid(V2) = 25 mL

Molarity of unknown acid (M2) = to be found = x

so, application of dimensional analysis yields the formula

MV1 -factor x V2 n {Where n-factor of diprotic acid is 2}

Plugging the values

0.1M X 10.7mL C 2 x 25ml

therefore x = unknown concentration = 0.0214 M

Trial 3

The volume of NaOH required for complete titration (V1) = 10.7 mL

Molarity of NaOH (M1) = 0.1 M

Volume of unknown acid(V2) = 25 mL

Molarity of unknown acid (M2) = to be found = x

so, application of dimensional analysis yeilds the formula

MV1 -factor x V2 n {Where n-factor of diprotic acid is 2}

Plugging the values

0.1M X 10.7mL C 2 x 25ml

therefore x = unknown concentration = 0.0214 M

So molarities for Trial 1 = 0.0214 M , Trial 2 = 0.0214 M , Trial 3 = 0.0214 M & Avg = 0.0214 M

Q 2)

Trial 4

The volume of NaOH required for complete titration (V1) = 10.7 mL

Molarity of NaOH (M1) = 1 M

Volume of unknown acid(V2) = 25 mL

Molarity of unknown acid (M2) = to be found = x

so, application of dimensional analysis yields the formula

MV1 -factor x V2 n {Where n-factor of diprotic acid is 2}

c = IM x 1.2mL 2 x 25mL

Therefore,for Trial 4, x = 0.024 M = molarity

The molarity determined by Trial 4 for appreciably matches with molarities determined by trials 1 to 3. ( small error of 0.0026 was observed). This suggests that the concentration of an unknown acid can be determined by both a small concentration of base and a high concentration of Base.

Q 3)

The volume required by 1 M NaOH was 1.2 mL whereas volume required by 0.1 M NaOH was 10.7 mL i.e volume of Base required for 1 M NaOH was far less than the volume of base required for 0.1 M NaOH.

The reason is that, no. of moles of the base in 1.2 mL NaOH = no. of moles of base in 10.7 mL 0.1 M NaOH

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