Solving a Buffer Problem in Chemistry: pH Calculation

In summary, the conversation is about a buffer problem in which the formula pH = pKa + log([B-]/[HB]) is used to calculate the pH of a buffer made up of 0.500 L each of .300 M KH2PO4 and .400 M K2HPO4. The question asks for the pH after the addition of 0.0500 mole of HCl and 0.0500 mole of NaOH, as well as the pH before the addition of any strong acid or strong base. The common ion effect is also mentioned, and it is suggested to figure out which component of the buffer is the base and use the new ratio in the equation to calculate the pH.
  • #1
gigi9
40
0
here is a buffer problem in my HW that I'm not sure how to do.
Plz help me. Plz show me how. Thanks a lot.
Formula: pH = pKa + log([B-]/[HB])
***A buffer is made up of 0.500 L each of .300 M KH2PO4 and .400 M K2HPO4. Calculate
a) The pH of the buffer after the addition of 0.0500 mole of HCl
b) The pH of the buffer after the addition of 0.0500 mole of NaOH.
 
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  • #2
Due to the common ion effect the concentrations of the buffer should be the same at equilibrium. Use the equation to calculate the pH before the addition of any strong acid or strong base.

The acid will react with the weak base. Figure out which component of the buffer is the base. After the mole to mole reaction, less base buffer component will be left and more of the conjugate acid will form. Write out an equation to see this for yourself. Use this new ratio and substitute into the equation.

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  • #3


Sure, I'd be happy to help you solve this buffer problem in chemistry. First, let's review some key concepts about buffers and pH calculations.

A buffer is a solution that resists changes in pH when small amounts of acid or base are added to it. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. In this case, we have a buffer made up of two components: KH2PO4, which is a weak acid, and K2HPO4, which is its conjugate base.

The pH of a solution is a measure of its acidity or basicity, and it is determined by the concentration of hydrogen ions (H+) in the solution. pH is measured on a scale from 0 to 14, where a pH of 7 is considered neutral, below 7 is acidic, and above 7 is basic. The lower the pH, the more acidic the solution is, and the higher the pH, the more basic the solution is.

Now, let's look at the formula provided: pH = pKa + log([B-]/[HB]). This is known as the Henderson-Hasselbalch equation and is used to calculate the pH of a buffer solution. In this equation, pKa is the negative logarithm of the acid dissociation constant, [B-] is the concentration of the conjugate base, and [HB] is the concentration of the weak acid.

a) To calculate the pH after the addition of 0.0500 mole of HCl, we first need to determine the new concentrations of the buffer components. Since 0.0500 mole of HCl is added, the concentration of KH2PO4 will decrease by 0.0500 M (since it is being consumed in the reaction) and the concentration of K2HPO4 will remain constant. This means that the new concentration of KH2PO4 will be 0.300 M - 0.0500 M = 0.250 M. We can then plug these values into the Henderson-Hasselbalch equation:

pH = pKa + log([B-]/[HB])
pH = 7.20 + log([0.400]/[0.250])
pH = 7.20 + log(1.60)
pH = 7.20 + 0.20
pH = 7.40

Therefore,
 

1. What is a buffer solution in chemistry?

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. It is typically composed of a weak acid and its conjugate base, or a weak base and its conjugate acid.

2. How do you calculate the pH of a buffer solution?

To calculate the pH of a buffer solution, you need to know the concentration of the weak acid or base, the concentration of its conjugate base or acid, and the acid dissociation constant (Ka) of the weak acid or base. You can then use the Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/[weak acid]).

3. What is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation is a mathematical relationship that describes the pH of a buffer solution. It is pH = pKa + log([conjugate base]/[weak acid]).

4. Can you provide an example of solving a buffer problem in chemistry?

Sure, let's say we have a buffer solution made of 0.1 M acetic acid and 0.1 M sodium acetate. The Ka of acetic acid is 1.8 x 10^-5. Using the Henderson-Hasselbalch equation, the pH of the buffer solution would be: pH = 4.74 + log(0.1/0.1) = 4.74. This means the buffer solution has a pH of 4.74 and is able to resist changes in pH when small amounts of acid or base are added.

5. What factors can affect the effectiveness of a buffer solution?

The effectiveness of a buffer solution can be affected by the concentrations of the weak acid and its conjugate base, the acid dissociation constant (Ka) of the weak acid, temperature, and the presence of other ions that can react with the weak acid or base. Additionally, dilution of the buffer solution can also decrease its buffering capacity.

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