Molarity Of Potassium Hydrogenphthalate Lab Report
Title : Preparation of a standard solution
Aim
To prepare a standard solution of potassium hydrogenphthalate which will later be used to standardize sodium hydroxide solution, NaOH.
Research question
How does different molarity of potassium hydrogen phthalate solution which is prepared by dilution process influence the volume of sodium hydroxide, NaOH used during titration process until the colourless phenolphthalein indicator solution turns pink where the volume of potassium hydrogenphthalate used is kept constant at (22.5 ± 0.01) cm³ throughout the process?
Independent variable : Concentration of potassium hydrogen phthalate, KHP.
Dependent variable : Volume of sodium hydroxide solution used at the end point of titration
Controlled …show more content…
Final volume of NaOH 33.00 34.0 35.0 45.00 44.0 43.00 55.00 56.00 58.0
Volume of NaOH used (± 0.1cm³ ) 67.00 66.00 65.00 55.00 56.00 57.00 45.00 44.00 42.00
Average volume of NaOH used, cm³
(Mean, xˉ) 66.00 56.00 43.70
Standard deviation
1.00
1.00
1.53
Table 4 shows quantitative data of Part C ( Titration )
Determine Precision of Data
Coefficient variation
(Standard deviation,s)/(mean,xˉ)x 100%
3.53/165.7 X 100%
= 2.13 %
Calculation of actual concentration of sodium hydroxide solution, NaOH
Actual concentration of sodium hydroxide solution used can be calculated by using the formula :
MaVa = A MbVb B
Where : M = Molarity; V = volume; a=acid; b = base; A = Number of moles of acid B = Number of moles of base
Balanced chemical equation
C8H5KO4(aq) + NaOH (aq) C8H4KO4Na (aq) + H2O
From equation, 1 mol of C8H5KO4 react with 1 mol of NaOH 1.When Ma = 0.5 mol/dm³,
((0.5 mol/dm³) ( 0.023 dm³ ))/(Mb ( 0.066 dm^3))= 1/1
Mb = MaVa/Vb
= 0.2 mol/dm³
2.When Ma = 0.4 mol/dm³,
((0.4 mol/dm³) ( 0.023 dm³ ))/(Mb ( 0.056 dm^3))= 1/1
Mb = MaVa/Vb
= 0.2 mol/dm³ 3.When Ma = 0.3 mol/dm³,
((0.3 mol/dm³) ( 0.023 dm³ ))/(Mb ( 0.044 dm^3))= 1/1
Mb = MaVa/Vb
= 0.2 mol/dm³ Hence, the concentration of NaOH is 0.2
Aim
To prepare a standard solution of potassium hydrogenphthalate which will later be used to standardize sodium hydroxide solution, NaOH.
Research question
How does different molarity of potassium hydrogen phthalate solution which is prepared by dilution process influence the volume of sodium hydroxide, NaOH used during titration process until the colourless phenolphthalein indicator solution turns pink where the volume of potassium hydrogenphthalate used is kept constant at (22.5 ± 0.01) cm³ throughout the process?
Independent variable : Concentration of potassium hydrogen phthalate, KHP.
Dependent variable : Volume of sodium hydroxide solution used at the end point of titration
Controlled …show more content…
Final volume of NaOH 33.00 34.0 35.0 45.00 44.0 43.00 55.00 56.00 58.0
Volume of NaOH used (± 0.1cm³ ) 67.00 66.00 65.00 55.00 56.00 57.00 45.00 44.00 42.00
Average volume of NaOH used, cm³
(Mean, xˉ) 66.00 56.00 43.70
Standard deviation
1.00
1.00
1.53
Table 4 shows quantitative data of Part C ( Titration )
Determine Precision of Data
Coefficient variation
(Standard deviation,s)/(mean,xˉ)x 100%
3.53/165.7 X 100%
= 2.13 %
Calculation of actual concentration of sodium hydroxide solution, NaOH
Actual concentration of sodium hydroxide solution used can be calculated by using the formula :
MaVa = A MbVb B
Where : M = Molarity; V = volume; a=acid; b = base; A = Number of moles of acid B = Number of moles of base
Balanced chemical equation
C8H5KO4(aq) + NaOH (aq) C8H4KO4Na (aq) + H2O
From equation, 1 mol of C8H5KO4 react with 1 mol of NaOH 1.When Ma = 0.5 mol/dm³,
((0.5 mol/dm³) ( 0.023 dm³ ))/(Mb ( 0.066 dm^3))= 1/1
Mb = MaVa/Vb
= 0.2 mol/dm³
2.When Ma = 0.4 mol/dm³,
((0.4 mol/dm³) ( 0.023 dm³ ))/(Mb ( 0.056 dm^3))= 1/1
Mb = MaVa/Vb
= 0.2 mol/dm³ 3.When Ma = 0.3 mol/dm³,
((0.3 mol/dm³) ( 0.023 dm³ ))/(Mb ( 0.044 dm^3))= 1/1
Mb = MaVa/Vb
= 0.2 mol/dm³ Hence, the concentration of NaOH is 0.2