Question

Henderson-Hasselback ( please solve using this equation!)

You need to make a 0.1M solution of Glycine (9.4), however your lab is completely out of glycine powder. you do however have 1 liter of 0.1M glycine pH 9 and 1 liter of 0.1M Glcine pH 10. How could you mix these two solutions togther to generate 200ml of your desired solution. pKA of glucine 9.25.

please outline all the steps as clearly as you can.

Answer 1

For these two solutions

1 liter of 0.1M glycine pH 9 and 1 liter of 0.1M Glycine pH 10

you have to understand that 0.1 M is the sum of the acid form (HA) and base form (A-).

Here HA is H2N-CH2-COOH or more exactly ⁺H3N-CH2-COO^-

and A^- is H2N-CH2-COO^-

Amino acid	pKa1	pKa2
Glycine	2.34	9.60

The value you provide for pKa = 9.25 is close to reported value of pKa2 = 9.60

You have to correct all calculations if you want to use 9.25

               

pH = pKa2 + log([A-]/[HA])         (HH equation)

log([A-]/[HA]) = pH –pKa2

[A-]/[HA] = 10^pH-pKa2

For the first solution

[A-]/[HA] = 10^9-9.6 = 10^-0.6 = 0.25 = 1/4

[A-] = (1/5)x0.1M = 0.02 M

[HA] = (4/5) x 0.1M = 0.08 M

For the second solution

[A-]/[HA] = 10^10-9.6 = 10^0.4 = 2.5 = 5/2

[A-] = (5/7)x0.1M = 0.071 M round it to 0.07

[HA] = (2/7) x 0.1M = 0.029 M round it to 0.03

For the new pH=9.4 buffer

[A-]/[HA] = 10^9.4-9.6 = 10^-0.2 = 0.63 = 2/3

[A-] = (2/5)x0.1M = 0.04 M

[HA] = (3/5) x 0.1M = 0.06 M

Mixing in any proportion the two 0.1 M solutions, will give a 0.1 M solution (HA+A^-)

Thus the question is reduced to finding the mix proportion of

[A-] = 0.02M and [A-] = 0.07M to give [A-] = 0.04M.

You may find a practical rule to calculate the proportion (maybe you know the rectangle rule).

Mix 3 parts solution pH=9 with 2 parts solution pH=10

The final volume being 200 mL:

Mix 120 mL solution pH=9 with 80 mL solution pH=10

Note:

If you mix 2 solutions:

   Volume V1 of solution 1, having concentration C1 (mol/L or % v/v or g/L)

   Volume V2 of solution 2, having concentration C2

The final solution will have volume V= V1+V2 and its concentration will be

    C = (C1V1 + C2V2) / (V1+V2) = (C1V1 + C2V2) / V

(If the concentrations are %w/w. replace all Volumes V by masses m).

A more difficult case is when you know C1, C2 , final C and final V and you have to find V1 and V2. Consider V1 = x as unknown and V2 = V-x.

CV = xC1 + (V-x)C2

    C = (C1V1 + C2V2) / (V1+V2)

CV1 + CV2 = C1V1 + C2V2

CV2 - C2V2= C1V1 - CV1

V2 (C-C2) = V1 (C1-C)

(C-C2)/ (C1-C) = V1/V2

You may write also

|C2-C| /|C1-C| = V1/V2

So, the ratio V1/V2 is easy to calculate. A mnemonic procedure to remember this formula is the use of a rectangle.

Write the value of C1 in the left high corner. Write the value of C2 in the left low corner. Draw diagonales: at their intersection (rectangle center) write the value of C.

Calculate along diagonales: |C1-C| and write the value in the right low corner; |C-C2| and write the value in the right high corner.

You have now on the right corner the proportion V1/V2 that is |C2-C| /|C1-C|.

C1                                    | C2-C|

                       C

C2                                    | C1-C|          (draw diagonales )

The proportion V1/V2 is equal with the proportion |C2-C| /|C1-C|.

How the procedure was applied here:

0.02M                             0.07-0.04=0.03

                     0.04M

0.07M                              |0.02-0.04|=0.02

The ratio V1/V2 is 0.03/0.02 = 3:2 = 3 parts/2 parts.

Total volume 200 mL is 5 parts (1 part = 40 mL)

Take 3x40 mL from solution 0.02M and 2x40 mL from solution 0.07 M.