Question

How do I determine the blank columns? KHP is potassium hydrogen phthalate. The amount of KHP was placed in a beaker mixed in DH2O and phenolphthalein and then 0.1M NaOH was titrated in until a color change was observed, indicating the endpoint.

Answer 1

NOTE - WHEN WE PERFORM TITRATIONS, THE BURETTE READINGS ARE ALWAYS IN mL NOT L. THAT IS THE MISTAKE IN YOUR TABLE.

• Standardization of NaOH ( finding the molarity) :

In the reaction of KHP (potassium hydrogen phthalate) with NaOH, one mole of KHP reacts with one mole of NaOH, as seen in the equation:

C₈H₄O₄H⁺ + OH^- ------ C₈H₄O₄^2- + H₂O

1. ​Concentration of KHP:

Given ( first row),

Mass of KHP = 0.7012 g.

Volume of NaOH = 33.37 mL.

= One mole of KHP ---- 204.32 g

x moles of KHP = 0.7012 g.

x = 3.431 x 10^-3 moles.

2. Concentration of NaOH:

= According to the equation,

One more of KHP reacts with ------ one mole NaOH

3.431 x 10^{-3 moles of KHP react with ------ x moles of NaOH}

x = 3.431 x 10^-3 moles of NaOH.

$= Molarity of NaOH = number of moles of NaOH \times \frac{1000}{volume}$

$= Molarity of NaOH = 3.431 \times 10-3 \times \frac{1000}{33.37}$

= Molarity of NaOH = 0.1028 M.

Using the same method, you can find the molarity of NaOH for the other corresponding masses of KPH and volume of NaOH. (same formulas, just plug in the different values)

After calculation you shoul get:

mass of KPH Volume of NaOH Molarity of NaOH

1. 0.7012 g 33.37 mL 0.1028 M

2. 0.7004 g 33.23 mL 0.1031 M

3. 0.7008 g 33.37 mL 0.1027 M

4. 0.7001 g 33.38 mL 0.1026 M

• Finding % KPH after titration :

​Given ( first row ),​

Mass of KPH = 1.0028 g

Volume of NaOH = 25.33 mL

Molarity of NaOH = 0.1 M

• Number of moles of NaOH = Molarity x Volume

Number of moles = 0.1 x 25.33 / 1000 = 2.533 x 10^-3

• From the above equation,

One mole of KPH react with ------ one mole of NaOH

x moles of KPH react with ------ 2.533 x 10^-3 moles of NaOH

x moles = 2.533 x 10^-3 moles of KPH.

• One mole of KPH ----- 204.32 g of KPH

2.533 x 10^-3 moles of KPH = x gm of KHP

x = 0.5175 g of KPH.

• The above mass of KPH found is the pure KPH present in the flask. The KPH weighed out before the titration is the impure sample. To find % KPH in the sample :

% purity = pure form / impure form x 100

= % purity of KPH = 0.5157 / 1.0028 x 100 = 51.6097 %.

Using the same method, you can find the % KPHfor the other corresponding masses of KPH and volume of NaOH. (same formulas, just plug in the different values)

After calculation you shoul get:

mass of KPH Volume of NaOH % KPH

1. 1.0028 g 25.33 mL 51.6097 %

2. 1.0014 g 25.31 mL 51.6410 %

3. 1.0013 g 25.28 mL 51.5623 %

4. 1.0006 g 25.29 mL 51.6187 %

Using these two tables, average and standard deviation has been calculated in the image.