Frequently Asked Question

The pH of the solution can be calculated using the following equation:
pH = pKa + log([A-] / [HA])
where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base (in this case, CH3SO3-), and [HA] is the concentration of the acid (in this case, CH3SO3H).

Given that the Ka of CH3SO3H is -1.9, we can calculate the pKa as follows:
pKa = -log(Ka) = -log(10^-1.9) = 1.9
We are given that the concentration of Cu(CH3SO3)2 is 0.1 M and the concentration of CH3SO3H is 5 M. Since Cu(CH3SO3)2 is a salt of a weak acid and a strong base, it will undergo hydrolysis to produce CH3SO3- and Cu2+ ions. The concentration of CH3SO3- produced by the hydrolysis of Cu(CH3SO3)2 can be calculated using the following equation:
[CH3SO3-] = Kw / [H+]
where Kw is the ion product constant for water (1.0 x 10^-14) and [H+] is the concentration of hydrogen ions in the solution. Since we do not know the concentration of hydrogen ions yet, we will have to use an iterative method to solve for it.

Assuming that the initial concentration of H+ ions is negligible, we can calculate the initial concentration of CH3SO3- as follows:
[CH3SO3-] = Kw / [H+] = 1.0 x 10^-14 / 10^-7 = 1.0 x 10^-7 M
We can then use this initial concentration of CH3SO3- to calculate the initial pH of the solution:
pH = pKa + log([CH3SO3-] / [CH3SO3H]) = 1.9 + log(1.0 x 10^-7 / 5) = 0.6
Now that we have an initial pH, we can use it to calculate a more accurate concentration of CH3SO3- produced by the hydrolysis of Cu(CH3SO3)2. We can then use this new concentration of CH3SO3- to calculate a more accurate pH. We can repeat this process until we get a consistent pH value.

After several iterations, we find that the pH of the solution converges to a value of 0.6. Therefore, the pH of 0.1M Cu(CH3SO3)2 + 5M CH3SO3H is approximately 0.6.