Question

I mean that R R. Po reacts with the same mercury (flows) from all the attempts made in the​

Answer 1

Answer:

Sulfuric acid is diprotic, the ionization of the first hydrogen contributes significantly to the pH. The ionization constant for the first hydrogen (Ka(1)) is very large resulting in 100% ionization of the first H+.

H2SO4 + H2O = H3O+ + (HSO4)1- Ka(1) = very large

The second Ka(2) is much smaller, so the presence of the H+ from the first ionization step will hinder significant ionization of the second hydrogen, so its contribution to pH can be considered using the equilibrium constant for the second ionization step.

(HSO4)1- + H2O = H3O+ + (SO4)2- Ka(2) = 1x10^-2

The first ionization step gives us a solution with a H3O+ concentration of 0.025 mol/L and an (HSO4)- concentration of 0.025 mol/L

Now, if you were so inclined, you could use an ICE table to do the equilibrium calculation for the ionization of the second hydrogen given an initial hydrogen ion concentration of 0.025 mol/L and the Ka(2) for H2SO4.

This gives you the following concentrations at equilibrium (2)

[(HSO4)-] = 0.025 - x mol/L

[H3O+] = 0.025 + x mol/L

[(SO4)2-] = x

When I plug these into the equation for Ka(2), I get an equation of x^2 + 0.035x -2.5x10^-4 =0

Using the quadratic equation to solve for x I get a value of 6.1x10^-3 mol/L

So, the resulting H3O+ concentration ends up being 0.025 + 0.0061 = 0.031 mol/L

pH = -log[H3O+] = 1.5

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