Design a buffer that has a pH of 6.70 using one of the weak acid/conjugate base

Question

Design a buffer that has a pH of 6.70 using one of the weak acid/conjugate base systems shown below.

HC2O4-

C2O42-

6.4×10-5

4.19

H2PO4-

HPO42-

6.2×10-8

7.21

HCO3-

CO32-

4.8×10-11

10.32

How many grams of the potassium salt of the weak acid must be combined with how many grams of the potassium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?

grams potassium salt of weak acid =  g

grams potassium salt of conjugate base =  g

Weak Acid Conjugate Base Ka pKa

HC2O4-

C2O42-

6.4×10-5

4.19

H2PO4-

HPO42-

6.2×10-8

7.21

HCO3-

CO32-

4.8×10-11

10.32

Explanation / Answer

With the Henderson-Hasselbalch equation :

pH = pKa + log [( HPO42-)/(H2PO4-)] (eq I)

and knowing that : ( HPO42-) +(H2PO4-) = 1M (eq II)

we can determinate de concentration of one them.sustituing pH and pKa :

6,7 = 7,21 + log [( HPO42-)/(H2PO4-)] ---> 10(6,7-7,21) =[( HPO42-)/(H2PO4-)] --> 0,309 =[( HPO42-)/(H2PO4-)]

--> 0,309 * (H2PO4-) = ( HPO42-) . (eq III)

Now sustituing the (eq III) in the (eq II) , we determinate the (H2PO4-) :

0,309 * (H2PO4-) +(H2PO4-) = 1M --> ( 0,309+1)*(H2PO4-) = 1M --> (H2PO4-) =0,76 M

Using the (eq III) we determinate the ( HPO42-) :

0,309 * (H2PO4-) = ( HPO42-) --> 0,309 * (0,76) = ( HPO42-) ---> ( HPO42-) = 0,23 M

Now with the MW of each compounds are calculated the grams :

MW(H2PO4-)= 96,64 g/mol ;MW(HPO42-)= 95,64 g/mol

W(H2PO4-) = 1L *(H2PO4-)*MW(H2PO4-) = 1L * (0,76mol/L)*96,64 g/mol = 73,45 g

W ( HPO42-) = 1L * ( HPO42-)*MW ( HPO42-)= 1L*(0,23mol/L)* 95,64 g/mol = 21,99 g

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