Calculate the pH of a 0.21 M CoCl3 solution. The Ka value for Co(H2O)63+ is 1.0 10-5.
Co(H2O)6^+3 ==> [Co(H2O)5(OH^-)]^+2 + H^+
Write the Ka expression.
Substitute x for H^+ and the Co compound on the right and 0.21-x for the Co compound on the left. Solve for x which will be (H^+), then convert to pH.
To calculate the pH of a solution, we need to determine whether the solute is acidic or basic. In this case, we are given the Ka value for the complex ion Co(H2O)63+ which represents the equilibrium between the hydrated cation and its conjugate base.
The given Ka value is 1.0 x 10^-5, which means it is a weak acid. Therefore, the hydrated cation, Co(H2O)63+, will act as a weak acid in solution, and its conjugate base, Co(H2O)62+, will act as a weak base.
To determine the pH, we need to calculate the concentration of H+ ions from the weak acid's dissociation. Since the expression for Ka is:
Ka = [H+][Co(H2O)62+]/[Co(H2O)63+]
We can assume that the Co(H2O)62+ concentration produced from the dissociation of Co(H2O)63+ is negligible compared to the initial concentration of Co(H2O)63+. This assumption allows us to simplify the equation to:
Ka = [H+]^2 / [Co(H2O)63+]
Rearranging the equation, we can isolate [H+]:
[H+]^2 = Ka * [Co(H2O)63+]
[H+] = sqrt(Ka * [Co(H2O)63+])
Now we can substitute the given values into the equation:
[H+] = sqrt((1.0 x 10^-5) * (0.21 M))
[H+] = 1.45 x 10^-3 M
Since the concentration of H+ ions will be the same as the concentration of OH- ions in a neutral solution, we can assume that [H+] = [OH-] in pure water. Therefore, the pH can be calculated using the equation:
pH = -log[H+]
pH = -log(1.45 x 10^-3)
pH = 2.84
So, the pH of a 0.21 M CoCl3 solution is approximately 2.84.