Assuming complete dissocaition, what is the pH of 4.93 mg/L BaOH2 solution

There is no BaOH2. You must mean Ba(OH)2

Ba(OH)2 ==> Ba^2+ + 2OH^-
4.93 mg/L = 0.00493 g/L.
mols/L = M = 0.00493/molar mass Ba(OH)2
1 mol Ba(OH)2 gives 2 mols OH^-
Therefore, M OH^- must be twice M Ba(OH)2.
Then pOH = -log(OH^-) and
pH + pOH = pKw = 14. You know pOH and pKw, solve for pH.

Ba(OH)2 is a strong electrolyte. Determine the concentration of each of the individual ions in a 0.800 M Ba(OH)2 solution.

(OH-): ??? M

To determine the pH of a solution, we need to know the concentration of the hydrogen ion (H+). In the case of a strong base like Ba(OH)2, complete dissociation occurs, which means it completely ionizes into Ba2+ and OH- ions. Since OH- is a hydroxide ion, we can use its concentration to calculate the pOH, and then convert it to pH.

First, let's convert the given concentration of Ba(OH)2 from mg/L to moles per liter (M):

1. Calculate the molar mass of Ba(OH)2:
Ba(OH)2 = 137.33 g/mol (Ba) + 2(16.00 g/mol (O) + 1.01 g/mol (H))
= 137.33 g/mol + 32.00 g/mol + 2.02 g/mol
= 171.35 g/mol

2. Convert the given concentration of 4.93 mg/L to moles per liter (M):
4.93 mg/L * (1 g / 1000 mg) * (1 mol / 171.35 g) = 2.88 × 10^-5 M

Now that we have the concentration of OH- ions, we can calculate the pOH:

pOH = -log10[OH-]
= -log10[2.88 × 10^-5]
= 4.54

Since the solution is basic, we can use the pOH to find the pH by subtracting it from 14:

pH = 14 - pOH
= 14 - 4.54
= 9.46

Therefore, the pH of the 4.93 mg/L Ba(OH)2 solution is 9.46.