Determining Ka by the half-titration of a weak acid Free Essay Example

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.

PH ±0.1

Qualitative observations

At ½ equivalence

5.0

When I recorded this, as we slowly added NaOH to the acid, there was a change of color from colorless to a very slight pink as the Phenolphthalein indicator changed color.

At equivalence

8.9

As I added the acetic acid to 250 cm3 of reaction mixture, there was no color change. Also as we measured the PH, the PH changed slowly but then changed very quickly at the solution approached equivalence. At this time, the indicator turned pink, when equivalence was reached

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:

Volume of NaOH (0.2%)

PH of solution (±0.2)

0

2.38

45

14

48

14

50

14

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.

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 NaOH, and then plotting a titration curve (method 2), we got a of 4.8±0.2%.

As I calculated both I can calculate the % error of both comparing it with the actual value, -4.76.[1]

% error of method 1= = 100 ±5%

% error of method 2= = 100 ±0.84%

The data I have concluded and summarized above is backed up by the data produced in the experiment and trends seen. We conclude that method 2 is more accurate as the % error is less and that our oringal method had ±5% error. This is clearly backed up in our %errors as 4.8 is much closer to the actual value 4.76. The data that supports our % errors is the graph.

It clearly shows a trend that as the volume of NaOH increased the PH rose, and the higher gradient signaling the equivalence point was at 28 cm3. Thus the graph clearly showed a half-equivalence point of 4.8 PH. Also the graph bolster that the PH at half-equivalence had to be less than 6, thus supporting the PH obtained by method 1, and hence the PKa obtained. Finally as for method 1, we simply recorded two results; we know that the datum that determined our percentage error was 5.0, and hence this data is what supports our %error.

Finally the confidence level for my conclusion is good. I got the results expected, as method 2 will always be more accurate than method 1 as the latter depends too much on qualitative and subjective recordings. Thus my confidence level for such conclusion is good. Also my confidence levels on the %error and PKa for method 1, is high as 5% error was small. Thus due to this low %error, my confidence level of the experiment done for method 1 is high. Even more the PKa obtained by method 2, has a higher confidence level as the % error was barely 0.84%. Thus method 2, has excellent confident level for its extremely low %error.

However the first factor that affects my confidence level is uncertainties. From the %error of PH, we got the %uncertainty of the PKa for method 1. Thus, we know, that from the total % error of 5%, 2% was made by systematic errors i.e uncertainties in this case. Thus the other 3% was caused by random error.

Similarly, for method 2, we got % uncertainties for the PH by the volume measure of NaOH. This %error was 4.2%, meaning 4.2% of the total error was caused by systematic error of the graph. Clearly this is bigger than the total %error of 0.84%. Thus this means that actually, even if our graph has on the y-axis an uncertainty of ±0.4, this is an over-estimate. This is since, while we can read a value off with this uncertainty, it can still be very close to the actual half-equivalence PH.

Thus this increases my confidence level, as it shows, that the systematic error of the graph y-axis uncertainty is very limited. Thus the biggest error is random error. This occurs when estimating the equivalence point from the titration graph, which is random error as it’s an estimate of the steepest point and hence has no uncertainty. Thus as we could underestimate or overestimate this value, it creates error, as we calculated the half equivalence from it. In this case, clearly we overestimated it as; the PKa from this method is higher than the actual one. Hence this error is directly reflected in our results limiting confidence levels.

Thus now we know what caused the % error for our methods. Hence, now my confidence level will increase as I know what type of error must be targeted to reduce most error. The random errors and systematic errors that constitute these percentages will be explained below, in the evaluation.

Evaluation:

From the results it is clear error was limited for method 1, 5%. We calculated that uncertainties make up at least 2% of that error. Thus systematic error only makes 2% of the error while random error makes 3% of the error.

Thus the significant error is random errors. This was due to the subjectiveness at seeing the half-titration points. As we relied on the fact that the phenolphatlein made the solution light pink, it was difficult to see such color change. Thus it was very easy to keep adding base, when there was already a color change. Hence our error was that we could overshoot the titration. As we added to much NaOH the color change seen was too much. So when we added the acetic acid, the PH at half-equivalence is higher so we overestimated the PKa. This was reflected directly on our results. Finally another less important random error was that pipettes leaked. Thus more NaOH was added. This while small also explains why we overestimated the PKa, as we overshooted the titration even more.

Finally our less significant errors were systematic error. They only make 2% of our errors. They were mainly caused by inaccuracy of our apparatus. The main systematic error caused was by the PH probe. The PH probe, first of all has great inaccuracy recording PH with a ±0.1 uncertainty. Thus as the PH recorded was small, the %uncertainty calculated is much bigger than it would be with a higher PH. The other uncertainty was caused by the inaccuracy of pipettes. When we measured the volume of the acetic acid, there was a systematic error as Burets have uncertainties of, ±0.10 cm3. Thus at a volume we measured of acetic acid at 25cm3, we had 0.4% error caused.

We can also analyze improvements for method 2. We used this method and generated it from data we had form method 1. However, the titration sketch clearly was much more accurate than method 1, as it yields 0.84% error of which 0.2% was caused by uncertainties. Thus as we got the results for the titration curve from method 1, the error that caused the systematic errors were the same. However the main cause of error is the random error. At calculating the equivalence point, we had to estimate the point with the maximum gradient. As this is subjective, there is human error. Hence, when we then halved that volume, we could overestimate or underestimate the error since we estimated the point with maximum gradient.

Improvements:

To reduce the random error firstly we must do more trials. Just by doing this, we will reduce the random error. Finally as the problem with the color change was that it was a qualitative observation. To improve this we can get a quantitative measurement. To do this we use a colorimeter. This is a device we will put behind the solution. This measures the exact absorbance or transmission of light. Thus as the light absorbance changes when there is a color change, when the colorimeter states such we know that the color change has occurred. Hence we know exactly the equivalence points. The significance of this improvement is that it would enable us to get qualitative results. Thus if the colorimeter very accurate we can decrease random error, as there is no human error. Also, as the colorimeter is accurate, systematic error will also be limited.

Another way we can improve is in the systematic errors. The first problem was measuring accurately volumes. As the pipettes had big uncertainties, the volume recorded had high %uncertainties. If we however use micropipettes, which have ±0.01 cm3 uncertainties, our volumes will be extremely accurate. Hence %uncertainties will be minimal. Also micropipettes allow much easier for drops of base to be dropped. Thus the significance of this improvement is that when we measure volumes, the equivalence point will occur, more exactly as we will be less likely to overshoot the solution.

Finally to solve the inaccurate measurements of PH we can get a PH sensor and data logger. These do real-time measurements and will state the PH with less uncertainty. It will also provide an alternative method for calculating the half-point. As the data logger draws the graph of the titration done, it can calculate the point with the highest gradient. Thus this will be the equivalence point. Hence we can calculate the PH at half the equivalence point of the graph as this is half the volume of base at equivalence. Thus clearly calculating a very accurate PH from the curve. The significance of this will be that it is a major improvement on method 2 and 1 as it is not qualitative. Thus it does not allow for human error. Hence as the sensor is also very accurate systematic error will also be limited as well as random error. Thus this method will get a very accurate PKa with low systematic and random errors.