Using the Henderson-Hasselbalch equation to find solution buffers.

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In summary, the problem involves producing 100.0 mL of 1.00 M potassium phosphate buffer solution of pH = 7.14, using 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and pure distilled H2O. The solution can be found by using the equations C(a) + C(b) = 100.0 mL and C(b) = C(a) x calculated ratio (0.85). The values for C(a) and C(b) can be obtained by solving the equation 7.14 = 7.21 + log [base]/[acid] and finding
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Homework Statement


As a technician in a large pharmaceutical research firm, you need to produce 100.0 mL of 1.00 M potassium phosphate buffer solution of pH = 7.14. The pKa of H2PO4- is 7.21.

You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O .

How much 1.00 M KH2PO4 will you need to make this solution?

C(a) represents the volume of the conjugate acid (in milliliters), and C(b) represents the volume of the conjugate base (in milliliters).

Homework Equations


C(a) + C(b) = 100.0 mL
C(b) / C(a) = calculated ratio (0.85)
C(b) = C(a) x calculated ratio (0.85)

C(a) + ( C(a) x (0.85) ) = 100.0 mL

C(a) = 100.0 / (1+(0.85)) mL

The Attempt at a Solution


I just for the life of me can't understand the section after the base/acid ratio is found.
7.14 = 7.21 + log [base]/[acid]
7.14 = 7.21 + log x
logx = pH - pKa (-.07)
x = 0.85 [base]/[acid] ratio

But after that I just honestly have no idea, I did a similar problem in class today but I just can't understand it now. I just would like to know what values belong where in the C(a) and C(b) equations and how to get those numbers so I can solve it myself. Thanks.
 
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  • #2
Winback said:
C(a) + C(b) = 100.0 mL

You add concentrations, yet you get volume as the answer? Correct this equation and you should be able to solve the problem.
 

Related to Using the Henderson-Hasselbalch equation to find solution buffers.

What is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation is a mathematical formula that describes the relationship between the pH of a solution, the pKa of an acid, and the concentration of the acid and its conjugate base.

How is the Henderson-Hasselbalch equation used to find solution buffers?

The Henderson-Hasselbalch equation can be used to determine the ratio of acid to conjugate base needed to create a buffer solution with a specific pH. By rearranging the equation and plugging in the desired pH and pKa values, the concentration of the acid and conjugate base can be calculated.

What are the assumptions made when using the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation assumes that the acid and its conjugate base are the only significant contributors to the solution's pH, and that the acid and conjugate base do not react with one another.

Can the Henderson-Hasselbalch equation be used for all acid-base systems?

No, the Henderson-Hasselbalch equation is only applicable to weak acids and their conjugate bases. Strong acids and bases, as well as polyprotic acids, cannot be analyzed using this equation.

How accurate is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation is a simplified model and may not be accurate for all solutions. Factors such as ionic strength, temperature, and activity coefficients can affect the accuracy of the equation. It is best used for initial estimations and may require further adjustments in practical applications.

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