posted by christine .
im trying to calculate the pH of a 1.0L soln containing 53.8 g of completely dissociated Mg(CN)2.
is this the diss. equation?
Mg(CN)2 + water --> MgOH + 2CN^-
i don't think i did the problem rite b/c i got 15.7 for the pH.
i got 53.8M Mg(CN)2 for the initial conc.
or i think it might be 2.03 = pH
i got this from the CN^- equilibrium concentration, which was 2(53.8)= 107.6
pH= -log (107.6)= 2.03
In the equilibrium concentration, you need that in moles per liter, NOT grams.
isnt the eq conc. 53.8 M (g/L)?