What is pH of the buffer 0.1 mol L-1 Na2HPO4/0.15 mol L-1 KH2PO4?

(Given Ka(H2PO4-)=6.2x10 power-8).

Sorry if the power is confusing. My computer does not have the symbol.

Thanks

To calculate the pH of a buffer solution, you need to consider the dissociation of the acid and the formation of its conjugate base. In this case, we have Na2HPO4, which is the conjugate base, and KH2PO4, which is the acid.

Let's start by writing down the dissociation reaction for KH2PO4:

KH2PO4 ⇌ H+ + K+ + HPO4^2-

The equilibrium constant for this reaction can be expressed as Ka = [H+][HPO4^2-]/[KH2PO4]. Given that Ka(H2PO4-) = 6.2x10^-8, we can use this value to calculate the concentration of H+ and HPO4^2- in the buffer solution.

Let's define x as the concentration of H+ and HPO4^2- in mol/L in the buffer solution. Since KH2PO4 and Na2HPO4 are in a 1:1 ratio, the concentration of KH2PO4 would be 0.15 - x, and the concentration of Na2HPO4 would be 0.1 - x.

Using the equilibrium constant expression, we have:

6.2x10^-8 = x^2 / (0.15 - x)

Since x is small compared to 0.15, we can simplify this equation further by ignoring x in the denominator:

6.2x10^-8 ≈ x^2 / 0.15

Rearranging the equation to isolate x^2, we have:

x^2 = 6.2x10^-8 * 0.15

x^2 ≈ 9.3x10^-9

Taking the square root of both sides, we find:

x ≈ 3.05x10^-4

Now that we have the concentration of H+, we can calculate the pH using the equation:

pH = -log [H+]

pH ≈ -log (3.05x10^-4)

pH ≈ 3.52

So, the pH of the buffer solution 0.1 mol L-1 Na2HPO4/0.15 mol L-1 KH2PO4 is approximately 3.52.

To determine the pH of the buffer solution, we can use the Henderson-Hasselbalch equation, which is given by:

pH = pKa + log ([A-]/[HA])

Where:
pH is the desired pH
pKa is the logarithmic acid dissociation constant of the species you're considering (in this case, H2PO4-)
[A-] is the concentration of the conjugate base (in this case, H2PO4-)
[HA] is the concentration of the weak acid (in this case, HPO42-)

In this case, the weak acid is Na2HPO4 and its conjugate base is HPO42-. The concentration of Na2HPO4 is 0.1 mol L-1, and the concentration of KH2PO4 is 0.15 mol L-1.

To determine the concentration of H2PO4- (HA), we first need to calculate the concentration of HPO42- (A-) using the equation:

[HPO42-] = [KH2PO4] / (1 + 10^(pKa - pH))

Given that the pKa value is 6.2x10^-8, we can calculate the concentration of HPO42- as follows:

[HPO42-] = (0.15 mol L-1) / (1 + 10^(-8- pH))

Next, we can substitute the concentrations into the Henderson-Hasselbalch equation and solve for the pH:

pH = -log(6.2x10^-8) + log([HPO42-] / [H2PO4-])

Substituting the values:

pH = -log(6.2x10^-8) + log(([KH2PO4] / (1 + 10^(-8- pH))) / [Na2HPO4])

Now, we need to solve this equation using an iterative method or a numerical solver to find the pH value.