# Buffer Calculation using the Henderson-Hasselbalch Equation

• jibjab
In summary, the conversation discusses the process of preparing 750 mL of a 0.5 M phosphate buffer with a pH of 6.7 and a pKa of 7.21. The use of the equation pH=pKa+log(B/A) is discussed, and the steps to solve for the amounts of NaH2PO4 and Na2HPO4 needed are explained, including properties of logarithms. The final step involves converting moles to mass.
jibjab

## Homework Statement

Prepare 750 mL of 0.5 M phosphate buffer, pH 6.7 and pKa 7.21. NaH2PO4 (119.96 g/mol) and Na2HPO4 (141.96 g/mol) is available to use.

pH=pKa+log(B/A)

## The Attempt at a Solution

I have the solution from my teacher, but I have no idea how he got the numbers. Could someone please explain how to use this equation? I'm trying to work backwards (I'm terrible at math, I learn the methods from solved examples) but I can't figure out where the numbers are coming from:

6.7=7.21 +log(B/A)
B=0.3090A (How did he get that?)

0.75L x 0.5 mol/L = 0.375 mol = B+A

1.3090A = 0.375 (Why is it 1.3090 and not 0.3090A like above?)

34.37g = 119.96 x (He put a symbol here that looks like <=) A=0.2865 mol
12.57g = 141.96 x <= B=0.885 mol

If someone would explain the process above, I would be extremely grateful. Thanks

jibjab said:
6.7=7.21 +log(B/A)
B=0.3090A (How did he get that?)

That's where you need to know the math that you don't know. I can show you how to do it, but if you don't know properties of logarithms it won't help you much:

$$6.7-7.21=-0.51=log\frac{B}{A}$$

Antilogarithm (both sides used as exponents):

$$10^{-0.51}=\frac{B}{A}$$

$$B=10^{-0.51}A=0.3090A$$

0.75L x 0.5 mol/L = 0.375 mol = B+A

1.3090A = 0.375 (Why is it 1.3090 and not 0.3090A like above?)

$$B+A=0.375$$

But we know B=0.3090A, so we can substitute it:

$$0.3090A + A = 1.3090A = 0.375$$

[/quote]34.37g = 119.96 x (He put a symbol here that looks like <=) A=0.2865 mol
12.57g = 141.96 x <= B=0.885 mol[/QUOTE]

Number of moles of A and B were calculated from the equations above, here it is just a conversion of moles to mass.

OHH I was confusing antilog with natural log. I did say I was terrible at math. Thanks for your help!

## 1. What is the Henderson-Hasselbalch equation and what does it calculate?

The Henderson-Hasselbalch equation is a mathematical formula used to calculate the pH of a buffer solution. It takes into account the concentration of the acid and its conjugate base, as well as the equilibrium constant of the acid dissociation reaction.

## 2. How is the Henderson-Hasselbalch equation derived?

The Henderson-Hasselbalch equation is derived from the equilibrium constant expression for the acid dissociation reaction, rearranged to solve for pH. It takes into account the relationship between the concentrations of the acid and its conjugate base at equilibrium.

## 3. What is the significance of the Henderson-Hasselbalch equation in biochemistry?

The Henderson-Hasselbalch equation is commonly used in biochemistry to calculate the pH of biological buffers, which are solutions that help maintain a stable pH in living organisms. It is also used to determine the optimal pH for enzyme activity and to predict the behavior of acid-base reactions in biological systems.

## 4. Can the Henderson-Hasselbalch equation be used for all buffer solutions?

No, the Henderson-Hasselbalch equation is only applicable to buffer solutions containing weak acids and their conjugate bases. It cannot be used for strong acids or bases, as their dissociation is complete and does not allow for the establishment of an equilibrium.

## 5. How accurate is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation is a simplified mathematical model that assumes ideal conditions, such as negligible ionic strength and no changes in pH due to dilution. It is most accurate for buffer solutions with a pH close to the pKa of the weak acid. However, in real-world situations, other factors may affect the accuracy of the equation and experimental measurements should be used for more precise calculations.

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