Question

Determine the quantity of each chemical that you are going to use for preparing the buffer: Calculate the amount of each chemich you will need to weigh out to make solutions A-F as listed in the "materials" section below. Also perform the follwing calculations:

1. How much of solution A and solution B would you need to mix to obtain 100ml of 0.2M phosphate buffer at pH 6.8 (relevant pKa is 7.2)? Will you need to add any extra water or just mix solutions A and B?

2. How much acid (5 and 1 N HCl solutions will be provided) will you need to add to a 1M Triz (pKa = 8.3) base solution to adjust the pH to the indicated pH of 8?

3. What hazards does imidazole present?

4. Why is the pKa of a buffer potentially dependent on temperature?

Materials

Distilled water, monobasic sodium phoshate, dibasic sodium phosphate, sodium chloride, imidazole, magnesium chloride, Tris base and (1M and 5M) HCl to make the following stock solutions:

A. 0.25L of 0.2M monobasic sodium phosphate

B. 0.25L of 0.2M dibasic sodium phosphate

C. 20mL of 2M NaCl

D. 20mL of 2M imidazole

E. 20mL of 0.5M MgCl2

F. 40mL of 1M Tris base

Answer 1

Buffer solution; The solution which resists the change in pH by the adition of strong acid or strong base is called Buffer solution.

Determination of quantity of each chemicak:

A: 0.25 L of 0.2 M monobasic sodium phosphate (NaH₂PO₄)

molarity=weight(g)/molecular weeight(g/mol)*1/volume(L)

weight=0.2 M*119.98 g/mol*0.25 L

=4.8 g.

B: 0.25L of 0.2 M of dibasic sodium phosphate

weight=molarity*molecularweight*volume

=0.2*141.96*0.25L

=7.09 g

Similarly the quantity of chemical is determined for remaining chemical solutions also.

Calculate the amount of each chemical will need to weigh as follows:

A:

Since, the quntity of solution A we got in the above calculations is 4.8 g, it is required to weigh is 4.8 g.

Therefore, the amount of chemical A is weighed is 4.8 g.

B:

Since the quantity of solution B we got is 7.09 g, it is required to weight chemical B is 7.09 g.

Therefore, the amount of chemical B weigh is 7.09 g.

Since, the quntity of each chemical is equal to the amount of each chemical, it is needed to weigh corresponding amount of each chemical.

Please folow the same procedure for remaing chemical solutions also.

1.

Total Molarity of Mixture of solution A and solution B is (0.2+0.2) M=0.4M.

Similarly, Total volume of the solution A and solution B is ((0.25+0.25) L=0.5L.

According to data given in the question (1), total molarity of mixture is 0.2 M and total volume is 100 mL.

Number of moes of above given data is,

Number of moles=Molarity*Volume

=0.2*100 mL*10^-3 L

=0.02 mol

The actual number of moles for the solution A and solution B is,

Number of moles= Molarity*volume

=0.4*0.5 mol

=2 mol

2 mols of mixture is prepared form the weighing of chemicals a and B is (4.8+7.09) g=11.89 g

0.02 moles of mixture is prepared from the weighing of chemicals A and B is,

=(0.02 mol*1.89 g)/2 mol

=11.89*10^-2 g.

Therefore, The amount of chemicals A and B added to prepare 0.2 M and 100 mL solution is 11.89*10^-2 g.

=