Determining Ka by the half-titration of a weak acid

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Determining Ka by the half-titration of a weak acid

To get the Ka of acetic acid, HC2H3O2 I will react it with sodium hydroxide. The point when our reaction is half-titrated can be used to determine the pKa. As I have added half as many moles of acetic , as NaOH, Thus, OH- will have reacted with only half of the acetic acid leaving a solution with equal moles of HC2H3O2 and C2H3O2-. Then I will use the Henderson-Hasselblach equation to get pKa.

CH3COOH + NaOH                    H2O + NaCH3COO

Results:

Below is a table that summarizes our results for the reaction of 1M of acetic acid with 1Molar of NaOH which 50cm3 was used. The table shows the PH record at ½ equivalence and at equivalence. We also recorded the observations we saw during the reaction.

Calculating the PKa

To calculate PKa, we will use the Henderson-Hasselbalch equation. Hence the calculations below show how using this we can calculate the PKa

 = PKa +

But at the half equivalence, the concentration of acetic acid and its salt ion are the same. Thus, we get:

 = PKa +

 = PKa

Now the PH was, so PKa= 5.0 ±0.1 = 5.0 ± 2%

5.0 ±2% =

 = 10-5 ±2%

Titration curve:

To get error we are going to sketch a titration curve, and from this measure the PH at half equivalence. To do this:

PH of acetic acid (1M):

Ka =

 = 10-4.76

 

 = (1×10-4.76)

So PH of acetic acid= 2.38

Now PH of NaOH, (1M)

Now concentration of NaOH, was 1M

So

 = 1

 

= -log(1) = 0±0.2%

So PH= 14±0.2%

Thus with these results we can plot this:

We know that at volume of NaOH of 45 and 48, the PH will still be 14 as it’s in excess by far, thus getting to the PH of NaOH as the PH measured

The PH of the solution has uncertainty of ±0.2, as this is the smallest division of our y-axis in our titration sketch.

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Now after plotting our titration results, we can see that the equivalence point the volume was at a volume was at 28cm3 as it has the steepest gradient. Thus, the half-equivalence is at half a volume, 14cm3. At this volume the PH is 4.8 ±0.2

Using this value, as

 = PKa +

 = PKa

PKa= 4.8 ±0.2

4.8±4.2% =

 = 10-4.8 ±4.2%

Conclusion:

I have concluded that the

 of acetic acid is -5±2% just using the data recorded (method 1). However from using that data and calculating the pH of acetic acid and ...

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