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
please help..
In the equilibrium concentration, you need that in moles per liter, NOT grams.
isnt the eq conc. 53.8 M (g/L)?
No, the equilibrium concentration should be in moles per liter (M). To find the equilibrium concentration, you need to convert the mass of Mg(CN)2 given (53.8 g) into moles.
First, calculate the molar mass of Mg(CN)2. The molar mass of Mg = 24.31 g/mol, C = 12.01 g/mol, and N = 14.01 g/mol. Multiply the atomic masses by the respective number of atoms and sum them up:
Molar mass of Mg(CN)2 = (24.31 g/mol * 1) + (12.01 g/mol * 1) + (14.01 g/mol * 2) = 95.24 g/mol.
Now, use the molar mass to convert grams to moles:
Moles of Mg(CN)2 = 53.8 g / 95.24 g/mol ≈ 0.565 mol.
Since the equation shows that 1 mole of Mg(CN)2 produces 2 moles of CN^-, the equilibrium concentration of CN^- ions will be twice the moles of Mg(CN)2:
Equilibrium concentration of CN^- ions = 2 * 0.565 mol / 1 L = 1.13 M.
To calculate the pH of the solution, you need to use the equation for the dissociation of water:
2H2O ⇌ H3O^+ + OH^-.
In pure water, the concentration of H3O^+ and OH^- ions are equal and equal to 1.0 x 10^-7 M. However, in your case, you have MgOH, which can hydrolyze to produce OH^- ions.
Since the concentration of OH^- ions is twice the concentration of CN^- ions in the equilibrium, the concentration of OH^- ions will be 2 * 1.13 M = 2.26 M.
Using the equation for pH:
pOH = -log[OH^-],
pOH = -log(2.26) = 0.65.
Since pH + pOH = 14, the pH of the solution will be:
pH = 14 - 0.65 = 13.35.
Therefore, the pH of the solution is approximately 13.35, not 15.7 or 2.03.