Assuming complete dissociation, what is the pH of a 2.55 mg/L Ba(OH)2 solution?
mols Ba(OH)2 = 0.00255g/molar mass = ?
mols (OH^-) = twice that.
M = mols/L.
To find the pH of a Ba(OH)2 solution, we need to understand the properties of the compound and the dissociation of its ions.
Ba(OH)2 is a strong base that dissociates fully in water according to the following reaction:
Ba(OH)2(s) → Ba2+(aq) + 2OH-(aq)
Since Ba(OH)2 is a strong base, it dissociates completely, meaning that for every mole of Ba(OH)2 that dissolves, we get one mole of Ba2+ ions and two moles of OH- ions.
First, we need to convert the concentration from mg/L to mol/L. To do this, we need the molar mass of Ba(OH)2.
The molar mass of Ba(OH)2 is:
Ba: 137.33 g/mol
O: 16.00 g/mol (there are two oxygen atoms)
H: 1.01 g/mol (there are two hydrogen atoms)
Adding these up gives us a molar mass of 137.33 + 16.00 + 1.01 + 1.01 = 155.35 g/mol.
Now we can convert 2.55 mg of Ba(OH)2 to moles:
2.55 mg/L × (1 g/1000 mg) × (1 mol/155.35 g) = 1.64 × 10^-5 mol/L
Since Ba(OH)2 dissociates to produce two OH- ions for every mole of Ba(OH)2, the concentration of OH- ions in the solution is twice the concentration of Ba(OH)2:
(2 × 1.64 × 10^-5 mol/L) = 3.28 × 10^-5 mol/L
Finally, we can determine the pOH of the solution. The pOH is the negative logarithm of the hydroxide ion concentration:
pOH = -log10(OH- concentration) = -log10(3.28 × 10^-5) ≈ 4.48
To find the pH, we can use the equation:
pH + pOH = 14
pH + 4.48 = 14
pH ≈ 14 - 4.48
pH ≈ 9.52
Therefore, assuming complete dissociation, the pH of a 2.55 mg/L Ba(OH)2 solution is approximately 9.52.