Just ignore the Cl^-.
Co(H2O)6Cl3 ==> Co(H2O)6^+3 + 3Cl^-
Then the hexaaquocobalt(III) ion ionizes as (and I will write this a funny way to help you see it).
Co(H2O)6^+3 ==>Co(H2O)5(H2O)^+3==> Co(H2O)5(OH)^+2 + H^+.
Then you write the Ka expression for that last part and solve for H^+ after doing an ICE chart.
i need help.
Calculate the pH of a 0.24 M CoCl3 solution. The Ka value for Co(H2O)63+ is 1.0 10-5.
how do u do these with like the Cl there?
1 answer
1 answer