The pH of human blood needs to be between 7.35 and 7.45. You want to prepare a b

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 H2PO4- and the conjugate base is HPO4^2- . You want the concentration of the acid to be 0.0100 M. 1. If the initial H2PO4- concentration is 0.0100 M, what is the initial concentration of HPO4^2- that will give a pH of 7.40? 2. What is the maximum molarity of acid that this buffer can neutralize without the pH dropping below 7.35? 3. What is the maximum molarity of base that this buffer can neutralize without the pH going above 7.45?

Explanation / Answer

According to Hendersson Hesselbatch equation, pH of buffer is

pH=pKa + log [Conjugate base or HPO42-]/ [Acid or H2PO4-]

7.40= 7.21+ log [H2PO4 2-]/[0.010]

7.40-7.21= log[H2PO4 2-]- log[0.10]

0.19= log[H2PO4 2-]- (-2)

2+0.19 = log[H2PO42-]

[H2PO4 2-] = antilog 2.19= 0.0155 M

2. pH=pKa+ log[HPO4 2-]/ [H2PO4-]

Let the molarity of acid added to be x

Then, 7.35= 7.21+log(0.0155-x)/ 0.010+x

7.35-7.21= log (0.0155-x)/ 0.010+x

0.14= log(0.0155-x)/ 0.010+x

antilog(0.14)= (0.0155-x)/ 0.010+x   (taking log on both sides)

1.38= (0.0155-x)/ 0.010+x

1.38(0.010+x)= 0.0155-x

or x= 7.13*10^-4

So, the maximum molarity of acid is 7.13*10^-4 M

3. 7.45= 7.21+log(0.0155+x)/(0.010-x)

7.45-7.21= 0.24= log 0.0155+x)/(0.010-x)

1.738= 0.0155+x/ 0.010-x

x= 6.86*10^-4 Mm

So, molarity of base is 6.86*10^-4

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