The pH of human blood needs to be between 7.35 and 7.45. You want to prepare a b
ID: 505209 • Letter: T
Question
The pH of human blood needs to be between 7.35 and 7.45. You want to prepare a buffer solution that gives a pH of 7.40. You decide to use a sodium phosphate buffer: the acid is H_2PO_4^- and the conjugate base is HPO_4^2-. You want the concentration of the acid to be 0.0100 M. If the initial H_2PO_4^- concentration is 0.0100 M, what is the initial concentration of HPO_4^2- that will give a pH of 7.40? What is the maximum molarity of acid that this buffer can neutralize without the pH dropping below 7.35? What is the maximum molarity of base that this buffer can neutralize without the pH going above 7.45?Explanation / Answer
A buffer system is a mixture of a weak acid or a weak base and its salt (conjugate base or conjugate acid, respectively) that permits solutions to resist large changes in pH upon the addition of small amounts of hydrogen ions (H+ ) or hydroxide ions (OH- ).It means a buffer helps maintain a near constant pH upon the addition of small amounts of H+ or OH- to a solution.
The Henderson-Hasselbalch equation provides a convenient way to think about buffers and pH:
pH = pKa + log [A-]/[HA]............(i)
pH = pKa + log [HPO42-]/ [H2PO4-] {pKa= 7.21 }
Q.1 given : pH = 7.40, [H2PO4-] = 0.01M , [HPO42-] = x
7.40 = 7.21 + log x/ 0.01
log x/0.01 = 0.19
x = 1.55 * 0.01 = 0.0155 M
Q.2 given pH = 7.35
Keeping initial mol of [H2PO4-] = 0.01
total moles of [HPO42-] = 0.01 (for ideal buffer [HA] = [A-]
Suppose X mol of acid is added to buffer.
7.35 = 7.21 + log [HPO42- + X] / [H2PO4- -X]
[HPO42- + X] / [H2PO4- -X] = 1.38
0.01 + X = 1.38 (0.01 -X) = 0.0138 -1.38X
1.38X +X = 0.0138- 0.01 = 0.0038
X= 0.0038 /2.38 = 0.0016
Maximum molarity of acid that this buffer can neutralize without pH dropping below 7.35 = 0.0016 M
Q.3 given pH = 7.45
Keeping initial mol of [H2PO4-] = 0.01
total moles of [HPO42-] = 0.01 (for ideal buffer [HA] = [A-]
Suppose X mol of base is added to buffer.
7.45 = 7.21 + log [HPO42- + X] / [H2PO4- - X]
[HPO42- + X] / [H2PO4- - X] = 1.74
0.01 + X = 1.74 (0.01 - X) = 0.0174 - 1.74X
1.74X +X = 0.0174- 0.01 = 0.0074
X= 0.0074 /2.74 = 0.0027
Maximum molarity of base that this buffer can neutralize without pH going above 7.45 = 0.0027 M
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