No of moles of OH-(aq) 1
The determination of the concentration of NaOH can be determined by the following equation:
Molarity of C8H5O4K × Volume of C8H5O4K = mol ratio of C8H5O4K
Molarity of NaOH ×Volume of NaOH mol ratio of NaOH
Molarity of C8H5O4K × Volume of C8H5O4K = 1
Molarity of NaOH ×Volume of NaOH 1
The volume Of NaOH can be determined by the titration result.
Results:
Calculations:
Molarity of the NaOH can be found by substituting the volume:
Molarity of C8H5O4K × Volume of C8H5O4K = 1
Molarity of NaOH ×Volume of NaOH 1
0.09529 mol dm3× 0.025 dm3 = 1
Molarity of NaOH × 0.02545 dm3
2.38×10 -3 = 1
Molarity of NaOH × 0.02545 dm3
Molarity of NaOH × 0.02545 dm3 =2.38 ×10 -3
Molarity of NaOH=0.0935166…
Molarity of NaOH~ 0.09mol dm3
Discussion:
In the calculation we assume that the mol ratio between the two solution is 1: 1
As the acid and alkali are both monobasic giving 1 mol of hydrogen ions and 1 hydroxide ions per mole.
In this experiment, the potassium hydrogen phthalate was standardized before it can be used to undergo titration. This is because the potassium hydrogen phthalate was in solid form and it can be found in pure state in solid form, thus it should be standardized as it is in solid form and titration involves both the base and acid to be in liquid form.
In the titration experiment phenolphthalein was used as an indicator in the experiment.
It is colourless in the presence of acid and pink in alkaline solution. The hydrogen ions in potassium hydrogen phthalate causes the phenolphthalein to become colourless. The hydrogen ions present in the potassium hydrogen phthalate will react with the hydroxide ions present in sodium hydroxide to from water. The presence of OH- causes the phenolphthalein to turn pink. Once the H+ and OH- reacts to form water, the H+ ions in the hydrogen phthalate will be used up to form water with the OH- ions as shown:
H+ (aq) + OH-(aq) → H2O
The excess OH- will turn the solution alkaline and thus turn to phenolphthalein slightly pink. This indicates complete reaction and determines the amount of NaOH solution used in the titration.
However if too much OH- is used it causes the phenolphthalein to turn dark pink, this disrupts the result as it cannot determine when the end point was reached.
Questions:
1.What effect would each of the errors described below have on the calculated value of the concentration of sodium hydroxide?
a) The burette is not rinsed with the sodium hydroxide
b) The pipette is not rinsed with the potassium hydrogen phthalate solution
c) The tip of the burette is not filled before titration begins
d) The conical flask contains some distilled water before the addition of potassium hydrogen phthalate
2. in using phenolphthalein as an indicator, we prefer to titrate from colourless to pink rather than pink to colourless. Suggest a reason for this.
3) why is it advisable to remove sodium hydroxide from the burette as soon as possible after the titration?
- In all the errors from a to b, the concentration of NaOH will be affected.
- the burette should be rinsed with NaOH as the burette was washed with distilled water. This causes the concentration of NaOH to become slightly dilute and causes the concentration.
- As the pipette was first washed with distilled water, if it is not rinsed with potassium hydrogen phthalate this causes further dilution to the solution and causes the concentration of the NaOH disrupted.
- The tip of the burette should be filled with NaOH as no t filling it will cause some air bubbles to form.
- This causes further dilution to potassium hydrogen phthalate and disrupts the concentration.
2. Because it is easier to check the end point of the experiment as it can be easy to check the end point once the solution turns from colourless to slightly pink.
3. If it's an 'older' burette with a glass tap then the sodium hydroxide can react with the glass jamming the tap. Modern PTFE taps do not suffer from this problem - but it is still good practice to wash out all apparatus after use anyway.
Conclusion:
The average volume of NaOH used is 25.45 cm3 and the concentration of NaOH used is 0.09 mol dm3
