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Obtaining PH Curves For Acidalkali Titrations Essay

Obtaining ph Curves For Acid alkali Titrations Essay Research Paper Obtaining pH curves for acid alkali titrations Planning In order to obtain a pH curve I must first devise a way in which to measure both the pH and the volume of alkali being.

Obtaining PH Curves For Acid/alkali Titrations Essay, Research Paper

Obtaining pH curves for acid/alkali titrations Planning ??????????? In order to

obtain a pH curve, I must first devise a way in which to measure both the pH

and the volume of alkali being added to the weak acid in order to draw a graph

from my results and then from this graph find Ka.Method ??????????? I will set

up the equipment as shown in the diagram below: ??????????? I will use

a pipette and pipette filer to take 25cm3 exactly of the weak acid

(0.1M ethanoic acid, CH3COOH) and place it in a beaker. I will then

take a burette and place in it 50cm3 of 0.1M sodium hydroxide

(NaOH).? I will then measure the exact

amount of NaOH added to the acid whilst using an accurate pH meter rather than

indicator to measure the change in pH. Before using the pH meter, I will first

dip it into two buffer solutions of pH 4 and pH 7 and calibrate it so that it

reads accurately. I will record the values for the pH and the volume of NaOH

added. Theoretically, this should give me a pH curve from which to work. In

order to obtain a value for Ka, I shall find the equivalence point (at around

25cm3 where the graph goes up vertically) and find the pH at this

point, I shall then divide this by two to give me the ½ equivalence point.? Having obtained this value, I can assume

that: [CH3COOH] = [CH3COO-] = [H3+O] And as: Ka = [CH3COO-] [H3+O] ??????????? ?? [CH3COOH] Therefore: Ka = [H3+O] Or: pKa = pH (at ½ equivalence point) Or: pH = -Log10 Ka Prediction ??????????? I know that

the curve for a weak acid/strong alkali should start higher up the axis than a

strong acid/strong base curve.? This

results in the vertical section of the graph being shorter and due to salt

hydrolysis, the curve will rise to 14 far quicker than it should. This is due

to the following mechanism: CH3COOH + NaOH → CH3COONa + H2O CH3COONa → CH3COO- + Na+ H2O→ H+ + OH- The H+ ions will then combine with the CH3COO-

ions to produce CH3COOH: H+ + CH3COO- → CH3COOH The removal of H+ ions leaves OH- ions

in excess.? This excess of dissociated

OH- ions leaves a strongly alkaline solution as the solution has the

composition of NaOH with Na+ and OH- ions in solution. ??????????? Although my

equivalence point should be at exactly 25cm3, this may not work out

experimentally due to variations in the concentrations of the acid and alkali

and inaccuracies associated with the use of a pH meter which may well be wrong

by the same amount every time, thus creating a set of results either higher or

lower than they should be.? Results?????????? Volume

(cm3) +- 0.1 pH

(1.d.p) +- 0.1 pH

(1.d.p) +- 0.1 Average

pH (2.d.p) +- 0.10 0 2.7 2.7 2.70 2 3.3 3.4 3.35 4 3.7 3.7 3.70 6 3.9 4.0 3.95 8 4.1 4.1 4.10 10 4.3 4.3 4.30 12 4.4 4.5 4.45 14 4.6 4.6 4.60 16 4.7 4.8 4.75 18 4.9 4.9 4.90 20 5.0 5.1 5.05 22 5.2 5.3 5.25 23 ?????????????? – 5.4 5.40 24 5.6 5.5 5.55 25 ?????????????? – 5.7 5.70 26 5.8 6.0 5.90 27 ?????????????? – 6.5 6.50 28 7.7 10.0 8.85 29 ?????????????? – 11.3 11.30 30 11.7 11.8 11.75 31 ?????????????? – 12.0 12.00 32 12.1 12.1 12.10 34 12.2 12.2 12.20 36 12.3 12.3 12.30 38 12.4 12.4 12.40 40 12.5 12.5 12.50 Calculation to find Ka Equivalence point = approximately 8.8 ½ equivalence point = approximately 4.4 At ½ equivalence point: [CH3COOH] = [CH3COO-] = [H3+O] As: Ka = [CH3COO-] [H3+O] ??????????? ?? [CH3COOH] Therefore: Ka = [H3+O] Or: pKa = pH (at ½ equivalence point) Or: pH = -Log10 Ka 4.4 = -Log10 Ka Ka = 10-4.4 Ka = 3.9 x 10-5Conclusion ??????????? As I

predicted the graph started higher up on the pH scale than that of a strong

acid/strong base titration, and the vertical section was shorter and it rose to

14 more quickly than it should due to salt hydrolysis. I was also correct in

assuming that the equivalence point may not be exactly 25cm3, as in

actual fact it was around 28cm3.?

This can be explained by errors in the pH meter, and slight variations

in the concentration of the solutions of CH3COOH and NaOH. I

calculated the acid constant (Ka) or the concentration of hydrogen ions [H3+O]

to be 3.9 x 10-5 that seems reasonable having checked the

calculation.Evaluation ??????????? Although I

have obtained reasonable results, I feel there are several ways in which my

results could be improved. The pH meter I used, despite taking quick readings

was not the most accurate possible.? If

I wanted to improve the overall accuracy of my results, I could use a more

accurate pH meter that remained accurate after calibration, a problem I

encountered with my pH meter. Furthermore, although I can blame my pH meter for

inaccuracy in my pH readings, it is more likely that the inaccuracy in the

equivalence point is due to slight differences in the concentration of the acid

and alkali.? This is a problem that is

difficult to remedy as I was provided with the solutions of 0.1M sodium

hydroxide and 0.1M ethanoic acid.?

However, if I made up these solutions accurately myself, I would have

more control over the accuracy of the concentration. If I eliminated these

errors, I think my experiment would be near perfect.? I could use phenolphthalein to show the exact point at which the

graph goes vertical using the colour change. This would be beneficial as it

would back up my pH readings and enable me to calculate the equivalence point

more easily.

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