- Level: International Baccalaureate
- Subject: Chemistry
- Document length: 1985 words
Determining Ka by the half-titration of a weak acid
Extracts from this essay...
Introduction
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. 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.
Middle
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 Â±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 itâs 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.
Conclusion
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. ________________ [1] IB chemistry data booklet pg 13
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