Write the balanced acid-base reactions and calculate the pH of the followingaqueous solutions:i) H2 S 0.01 M (Ka1 (H2 S) = 9.5×10-8 ; Ka2 (HS-) = 1.3×10-13 )

The first step is to write the balanced acid-base reactions for H2S.

H2S is a diprotic acid, meaning it can donate two protons (H+). The two ionization steps are:

1. H2S ⇌ H+ + HS- (Ka1 = 9.5×10^-8)
2. HS- ⇌ H+ + S^2- (Ka2 = 1.3×10^-13)

Next, we calculate the pH of the 0.01 M H2S solution.

For the first ionization step, we can set up an ICE (Initial, Change, Equilibrium) table:

Where x is the concentration of H+ ions, which we want to find.

We can then set up the expression for Ka1:

Ka1 = [H+][HS-]/[H2S] = (x)(x)/(0.01-x) = 9.5×10^-8

Assuming x is much smaller than 0.01 (which is a reasonable assumption given the small value of Ka1), we can simplify this to:

x^2/0.01 = 9.5×10^-8

Solving for x gives x = sqrt(9.5×10^-8 * 0.01) = 9.74×10^-6

The pH is then -log[H+] = -log(9.74×10^-6) = 5.01

For the second ionization step, the concentration of HS- is approximately equal to x, and the concentration of H+ and S^2- are both approximately 0. However, because Ka2 is so small, we can assume that the second ionization step does not contribute significantly to the pH of the solution. Therefore, the pH of the 0.01 M H2S solution is approximately 5.01.