As a technician in a large pharmaceutical research firm, you need to produce 400
ID: 571163 • Letter: A
Question
As a technician in a large pharmaceutical research firm, you need to produce 400. { m mL} of 1.00 M potassium phosphate buffer solution of exttip{ m pH}{pH} = 7.08. The { m p}K_{ m a} of m {H_2PO_4}^- is 7.21. You have the following supplies: 2.00 m L of 1.00 M~ m KH_2PO_4 stock solution, 1.50 m L of 1.00 M~ m K_2HPO_4 stock solution, and a carboy of pure distilled m H_2O. How much 1.00 M~ m KH_2PO_4 will you need to make this solution? Express your answer to three significant digits with the appropriate units.
Explanation / Answer
The way you have submitted the question makes it difficult to read the question. However, what I could understand is that you want to prepare potassium phosphate buffer solution.
The ionization of dihydrogen phosphate can be written as
H2PO4-(aq) -------> H+(aq) + HPO42-(aq); pKa = 7.21
Use the Henderson-Hasslebach equation to find out the ratio of the acid and its conjugate base as
pH = pKa + log [HPO42-]/[H2PO4-]
====> 7.08 = 7.21 + log [HPO42-]/[H2PO4-]
====> log [HPO42-]/[H2PO4-] = 7.08 – 7.21 = -0.13
====> [HPO42-]/[H2PO4-] = antilog (-0.13) = 0.7413
====> [HPO42-] = 0.7413*[H2PO4-] ……(1)
Again, the total phosphate concentration is given as 1.00 M; therefore, we have
[HPO42-] + [H2PO4-] = 1.00 M
====> 0.7413*[H2PO4-] + [H2PO4-] = 1.00 M
====> 1.7413*[H2PO4-] = 1.00 M
====> [H2PO4-] = (1.00 M)/(1.7413) = 0.5743 M.
[HPO42-] = (1.00 M) – (0.5743 M) = 0.4257 M.
Next use the dilution equation to find out the volumes of the stock solutions required. We prepare 400 mL of the buffer and we have 2.00 L of 1.00 M KH2PO4 and 1.50 L of 1.00 M K2HPO4.
Use the dilution equation: M1*V1 = M2*V2
where M1 = concentration of the supplied stock solution; V1 = volume of the stock solution taken; M2 = concentration of the final solution and V2 = volume of the buffer solution.
Plug in values and obtain
KH2PO4
(1.00 M)*V1 = (0.5743 M)*(400 mL)
====> V1 = (0.5743 M)*(400 mL)/(1.00 M) = 229.72 mL.
KHPO4
(1.00 M)*V1 = (0.4257 M)*(400 mL)
====> V1 = (0.4257 M)*(400 mL)/(1.00 M) = 170.28 mL.
We shall add 229.72 mL of 1.00 M KH2PO4 and 170.28 mL of 1.00 M K2HPO4 to prepare 400 mL of 1.00 M potassium phosphate buffer.
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