Please show me where/if I made a mistake in the following calculations. Backgrou

Question

Please show me where/if I made a mistake in the following calculations.

Background info, NaOH is .02M. The EP's were found by looking at a graph; I believe they are reasonably accuate. I am not sure about the Ka part, and the concentration calculations are likely way wrong.

1a. Ka of Citric Acid

Equivalence Point 3 = 60 mL, pH 8.3
Half Equivalence Point 3 = 30 mL, pH 5.0
                  Ka = 10^-5.0 = 1.0x10^-5

Equivalence Point 2 = 40 mL, pH 5.5
Half Equivalence Point 2 = 20 mL, pH 4.o
Ka = 10^-4 = 1.0x10^-4

Equivalence Point 1 = 20 mL, pH 4.0
Half Equivalence Point 1 = 10 mL, pH 3.3
                  Ka = 10^-3.3 = 5.0x10^-4

1b. Concentration of Citric Acid

Moles of acid = moles of NaOH added at the equivalence point:
                  .02 M NaOH x .0300 L NaOH added = .0006 moles NaOH at the equivalence point

This means there were .0006 moles of Citric acid at the start of the reaction
                  Concentration of Citric Acid = moles/ initial volume (L ) = .0006 moles/ .02 L = .03 M

2a. Ka of Phosphoric Acid

For the phosphoric acid, there are 2 visible equivalence points. The third one would be above pH 13, so we do not attempt finding the third Ka. From the graph, we determine the E.P., the divifding to find the

Explanation / Answer

1b. you are wrong at calculating number of moles of citric acid.

And in calculating number of moles of NaOH, we have to take volume of NaOH at equivalence point that is 60ml. (not 30ml).

Citric acid is a triprotic acid, and we will be titrating all three protons. We know the average molarity of the NaOH. we also know the volume of NaOH required to reach the endpoint in each titration.

Since molarity * volume (in liters!) = mol NaOH, we know how many moles of NaOH were delivered to neutralize the acid. SInce the acid is triprotic, we convert mol NaOH --> mol citric acid using the balanced reaction above:

mol NaOH at equivalence point * ( 1 mol citric acid / 3 mol NaOH) = mol citric acid

.02 M NaOH x .0600 L NaOH added = .0012 moles NaOH at the equivalence point

This means there were .0012/3 moles of Citric acid at the start of the reaction=0.0004 moles of citric acid

To calculate the molarity of the citric acid, divide mol citric acid by the volume of the acid titrated

  Concentration of Citric Acid = moles/ initial volume (L ) = .0004 moles/ .02 L = 0.02 M

2a) H3PO4 is a  triprotic acid, and we will be titrating all three protons. But here you have only two titrations

therefore we calculate number of moles of H3PO4 after second equivalence point.

Here you were wrong at calculating number of moles of NaOH by taking volume at half equivalence point. we have to take volume at equivalence point.And no. of moles of Phosphoric acid = no. of moles of NaOH/2.

Moles of acid = moles of NaOH added at the equivalence point/2 (since only two titrations)
.02 M NaOH x 0.045 L NaOH added = .0009 moles NaOH at the equivalence point

This means there were .0009/2 =0.00045 moles of Phosphoric acid at the start of the reaction. To calculate the molarity of the Phosphoric acid, divide mol Phosphoric acid by the volume of the acid titrated

Concentration of Phosphoric Acid = moles/ initial volume (L ) = 0.00045 moles / .02 L = .0225 M

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