Question

Preparation of Phosphate Buffer Rxn:

Purpose:

The purpose of lab this week is to prepare a 0.05M sodium phosphate buffer, use a pH meter to adjust the pH of this buffer, and to calculate theoretical pH changes upon addition of acid/ base. Your theory will then be correlated against your actual observational pH changes.

Solutions to be made

Molecular Weight Table

Solution	Volume
1.0M HCL	10ML
1.0 M NaOH	20ml
0.05M Sodium Phosphate: *?g NaH2PO4 H2O + *?g Na2HPO4 7H2O, pH 7.0 (buffer)	100ml

*Do the calculation of grams needed

Chemical	Molecular Weight or Stock Concentration
NaH2PO4 H2O (Monosodium phosphate, monohydrate)	137.99 g/mol
Na2HPO4 7H2O (Disodium phosphate, heptahydrate)	268.07 g/mol
NaOH	40.00 g/mol
HCl	36.46 g/mol

Notes :

1. Prepare the solution to ~25-50 ml less than the final volume when you anticipate adjusting the pH of the solution to allow volume for adding the acid or base. This will allow some wiggle room for adding the acid or base and won’t cause changes to your overall concentration. Once the pH is correct, add more water to bring the solution up to the final volume. 

2. Measure volumes in graduated cylinders; volumes marked on beakers and bottles are inaccurate. 

3. Stir bars are valuable and love to go down the drain. Remove your stir bar and rinse it separately from your beaker. 


Calculation

4. Do the following calculation: to 50 ml of sodium phosphate buffer at pH 7.0 is added 2 mL of 0.5M NaOH. Calculate the resulting pH change and show all of your working

5. (ii) Now actually add the 2 ml of 0.5M NaOH to 50 ml of your buffer solution and stir carefully. Measure the new pH. Compare the experimental with the theoretical result. Is the sodium phosphate still acting as a buffer? 


   Preparation of Phosphate Buffer Rxn:

   Purpose:

   The purpose of lab this week is to prepare a 0.05M sodium phosphate buffer, use a pH meter to adjust the pH of this buffer, and to calculate theoretical pH changes upon addition of acid/ base. Your theory will then be correlated against your actual observational pH changes.

   Solutions to be made

   Molecular Weight Table

   Solution	Volume
   1.0M HCL	10ML
   1.0 M NaOH	20ml
   0.05M Sodium Phosphate: *?g NaH2PO4 H2O + *?g Na2HPO4 7H2O, pH 7.0 (buffer)	100ml

   *Do the calculation of grams needed

   Chemical	Molecular Weight or Stock Concentration
   NaH2PO4 H2O (Monosodium phosphate, monohydrate)	137.99 g/mol
   Na2HPO4 7H2O (Disodium phosphate, heptahydrate)	268.07 g/mol
   NaOH	40.00 g/mol
   HCl	36.46 g/mol

   Notes :

6. Prepare the solution to ~25-50 ml less than the final volume when you anticipate adjusting the pH of the solution to allow volume for adding the acid or base. This will allow some wiggle room for adding the acid or base and won’t cause changes to your overall concentration. Once the pH is correct, add more water to bring the solution up to the final volume. 

7. Measure volumes in graduated cylinders; volumes marked on beakers and bottles are inaccurate. 

8. Stir bars are valuable and love to go down the drain. Remove your stir bar and rinse it separately from your beaker. 


9. Calculation

10. Do the following calculation: to 50 ml of sodium phosphate buffer at pH 7.0 is added 2 mL of 0.5M NaOH. Calculate the resulting pH change and show all of your working

11. (ii) Now actually add the 2 ml of 0.5M NaOH to 50 ml of your buffer solution and stir carefully. Measure the new pH. Compare the experimental with the theoretical result. Is the sodium phosphate still acting as a buffer? 


Please answer all of the QUESTIONS and do the calculation needed to start the lab. Don't just give the formula show the work also to better understand.

Answer 1

According to the Henderson-Hasselbulch equation:

pH = pKa2 + Log{n(Na2HPO4.7H2O)/n(NaH2PO4.H2O)}

i.e. 7 = 7.2 + Log{n(Na2HPO4.7H2O)/n(NaH2PO4.H2O)}

i.e. Log{n(NaH2PO4.H2O)/n(Na2HPO4.7H2O)} = 0.2

i.e. n(NaH2PO4.H2O)/n(Na2HPO4.7H2O) = 10^0.2 = 1.585

Where 'n' is the corresponding no. of millimoles

Here, n(NaH2PO4.H2O) + n(Na2HPO4.7H2O) = 0.05 mmol/mL * 100 mL = 5 mmol

i.e. n(NaH2PO4.H2O) = 5 - n(Na2HPO4.7H2O)

Therefore, {5 - n(Na2HPO4.7H2O)}/n(Na2HPO4.7H2O) = 1.585

i.e. n(Na2HPO4.7H2O) = 5/2.585 = 1.934 mmol

Hence, the mass of Na2HPO4.7H2O = 1.934 mmol * 268.07 g/mol = 0.518 g

Now, n(NaH2PO4.H2O) = 5 - 1.934 = 3.066 mmol

Hence, the mass of NaH2PO4.H2O = 3.066 mmol * 137.99 g/mol = 0.423 g

Addition of acid:

Millimoles of acid = 1 mmol/mL * 10 mL = 10 mmol

millimoles of HCl remaining = 10 - 3.066 = 6.934 mmol

i.e. [HCl] = [H⁺] = 6.934 mmol/(100+10) mL = 0.06304 M

Therefore, pH = -Log[H⁺] = -Log(0.06304) = 1.2

The change in pH = 1.2 - 7 = -5.8

Addition of base:

Millimoles of base = 1 mmol/mL * 20 mL = 20 mmol

millimoles of NaOH remaining = 20 - 1.934 = 18.066 mmol

i.e. [NaOH] = [OH^-] = 18.066 mmol/(100+20) mL = 0.15055 M

Therefore, pH = 14 - (-Log[OH^-]) = 14 - (-Log0.15055) = 14-0.82 = 13.18

The change in pH = 13.18 - 7 = 6.18