Calculate the pH of a buffer solution prepared by mixing equal volumes of 0.20 M NaHCO3 and

0.10 M Na2CO3. (Ka = 4.3 x10-7)

What's wrong with substituting the numbers into the Henderson-Hasselbalch equation? HCO3^- is the acid ad CO3^2- is the base.

To calculate the pH of a buffer solution, we need to consider the acid-base equilibrium in the solution. In this case, NaHCO3 and Na2CO3 are both salts of weak acids (H2CO3 and HCO3-), and their conjugate bases.

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:

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

Where pH is the desired pH, pKa is the logarithm of the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

In this case, we have equal volumes of 0.20 M NaHCO3 and 0.10 M Na2CO3. Since the volumes are equal, the concentrations of their conjugate base and weak acid will also be equal.

First, we need to determine the concentration of the weak acid and the conjugate base:

[HCO3-] = [NaHCO3] = 0.20 M
[CO32-] = [Na2CO3] = 0.10 M

Next, we need to calculate the pKa of H2CO3 using the given Ka value:

Ka = [H+][CO32-]/[HCO3-]

Since [H+]/[HCO3-] is the same as [HCO3-]/[H2CO3], we can rearrange the equation as:

Ka = [HCO3-][HCO3-]/[H2CO3]
[H2CO3] = [HCO3-][HCO3-]/Ka

Now, substitute the known values into the equation:

[H2CO3] = (0.20 M)(0.20 M)/4.3 x 10^-7

Simplify the equation by calculating the numerator and dividing it by the denominator:

[H2CO3] = 0.0400 M^2 / 4.3 x 10^-7
[H2CO3] = 93.02 M

Since [HA] = [H2CO3], [HA] = 93.02 M

Now, we can calculate the pH using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])
pH = log(4.3 x 10^-7) + log(0.10 M / 93.02 M)

Calculate the logarithms and divide the concentrations:

pH = -6.37 + log(0.001074)

Use a calculator to determine the value of log(0.001074), then add it to -6.37 to get the final pH value.

P.S. Please note that I have rounded the numbers to make the calculations simpler, but it's always best to use the exact values for more accurate results.