# The Ka for the H2PO4–/HPO42– buffer is given by the equation:

Ka= ([HPO4^-2]gamma [H^+])/[H2PO4^-]=10^-7.20

What is the pH of a 2:3 mixture of H2PO4– to HPO42– at 0.10 M ionic strength? (See the table for the activity coefficients.)

I just am not sure how to figure out the concentrations.. I know I have to solve for H+ but I am not sure what to put for HPO4^-2 and H2PO4- Thank you!

## I must confess that I, too, am confused but mostly because I don't know what gamma is in your equation. I THINK, but I'm not positive, that you look in the table, find the activity coefficient for H2PO4^- and HPO4^2-. Then I would be inclined to use 2 for H2PO4^- and 3 for HPO4^2- and multiply each by its activity coefficient. Since it's the ratio in Ka it makes little difference what REAL concn it is because 2M/3M is the same as 0.2/0.3 or 0.002/0.003. And since you are given the ionic strength that will correct for whatever difference you have because the activity coefficient will not be the same for each. I think this takes care of the 2:3 mixture of 0.1 M ionic strength part of the problem if you know what to do with gamma.

## Ah, the wonderful world of buffers and pH calculations! So, you have a 2:3 mixture of H2PO4– to HPO42–, and you want to find the pH at a certain ionic strength. Let me help you out!

First, let's assign some variables so we can keep track of everything:

[H2PO4–] = x (concentration of H2PO4–)

[HPO42–] = y (concentration of HPO42–)

Now, we need to relate x and y using the given ratio of 2:3. If we assume the total volume of the mixture is V, then we can say:

x = (2/5)V

y = (3/5)V

Next, we need to find the actual concentrations of H2PO4– and HPO42–. This is where the ionic strength comes into play. Let's assume the activity coefficients for H2PO4– and HPO42– are γ1 and γ2, respectively. Then,

[H2PO4–]actual = x * γ1

[HPO42–]actual = y * γ2

Now, we can substitute these values into the Ka expression you provided:

Ka = ([HPO42–]actual * [H+]) / [H2PO4–]actual = 10^(-7.20)

But wait, we're missing [H+], which is exactly what we're trying to find! To solve for it, we need additional information. Is there any additional information about the system or any equilibrium conditions given?

Remember, I'm here to help with humor, not just calculations! So feel free to ask me anything else or if you need help with any other part of the problem.

## To solve this problem, let's assign some variables:

Let x = [H2PO4^-] (concentration of H2PO4^-)

Let y = [HPO4^2-] (concentration of HPO4^2-)

Since we are given a 2:3 mixture of H2PO4^- to HPO4^2-, we can write:

x = (2/5)total concentration (1)

y = (3/5)total concentration (2)

We also have the equilibrium constant expression for the buffer:

Ka = ([HPO4^2-]γ[H+])/[H2PO4^-]

Given that Ka = 10^-7.20, and assuming the activity coefficients (γ) for H2PO4^- and HPO4^2- are both equal to 1, we can substitute the values into the equation:

10^-7.20 = (3/5)x(γ[H+]) / (2/5)x

Since the ionic strength is 0.10 M, we need to consider the effect of ionic strength on the activity coefficients. In this case, we are given a table with activity coefficients. To account for the ionic strength, we can write the expression as follows:

10^-7.20 = (3/5)x(γ[H+]) / (2/5)x * I * √(1 + (I/2K)) (where I is the ionic strength and K is a constant related to the medium)

Given that I = 0.10 M, we can substitute it into the equation:

10^-7.20 = (3/5)x(γ[H+]) / (2/5)x * (0.10) * √(1 + (0.10/2K))

To solve for the pH, we need to solve for [H+]. To do this, we rearrange the equation:

[H+] = 10^( -7.20 ) * (2/3) * x / (γ * (0.10) * √(1 + (0.10/2K)))

To find the pH, we need to take the negative logarithm (pH = -log10[H+]) of the concentration:

pH = -log10( 10^( -7.20 ) * (2/3) * x / (γ * (0.10) * √(1 + (0.10/2K))) )

This equation will give you the pH of the solution. Note that to solve fully, you would need to know the value of the constant K, which is not provided in the question.

## To solve for the pH of a 2:3 mixture of H2PO4– to HPO42– at a 0.10 M ionic strength, you need to consider the dissociation equilibrium of the phosphoric acid, H3PO4.

First, let's assume the initial concentration of H2PO4– is x, and the initial concentration of HPO42– is y. Since the mixture is in a 2:3 ratio, we can write the equations:

H2PO4– → HPO42– + H+

2x → y + H+

Now, you'll need information on the dissociation equilibrium to calculate the concentrations of H2PO4–, HPO42–, and H+.

Given in the question:

Ka = ([HPO4^-2]γ[H^+])/[H2PO4^-] = 10^-7.20

In this equation, [HPO42–] is the concentration of HPO42–, and [H2PO4–] is the concentration of H2PO4–. The γ indicates the activity coefficient, which you can find in the given table.

After that, you'll also need to consider the ionic strength (I) of the solution, which can be calculated using the formula:

I = 1/2 * (2 x [H2PO4–]γ + 3 x [HPO42–]γ)

Given in the question:

I = 0.10 M

Now, you have the necessary equations and information to solve for the concentrations of H2PO4–, HPO42–, and H+.

1. Start by calculating the activity coefficients using the given table.

2. Use the activity coefficients to write out the equation for the ionic strength (I).

3. Plug the calculated activity coefficients and ionic strength into the Ka equation.

4. Rearrange the Ka equation to solve for [H+].

5. Use the equation for H2PO4– and HPO42– ratios (2:3) to write expressions for [H2PO4–] and [HPO42–].

6. Substitute the expressions for [H2PO4–], [HPO42–], and [H+] into the equation for the dissociation of H2PO4–.

7. Solve the equation to obtain the value of [H+].

8. Take the negative logarithm (pH) of the [H+] value to find the pH of the mixture.

Remember to keep track of units and ensure that the values used in calculations are in proper scientific notation.