Find the pH of a solution containing 0.344 g of barium hydroxide in 620 mL of water.
This is a base. That means the M you find will be the [OH-]
1. Calculate the molar mass of barium hydroxide (Ba(OH)2):
Ba: 137.33 g/mol
O: 16.00 g/mol x 2 = 32.00 g/mol
H: 1.01 g/mol x 2 = 2.02 g/mol
Molar mass of Ba(OH)2 = 137.33 + 32.00 + 2.02 = 171.35 g/mol
2. Calculate the moles of barium hydroxide:
0.344 g / 171.35 g/mol = 0.002 g
3. Calculate the molarity of barium hydroxide in the solution:
Volume of solution = 620 mL = 0.62 L
Molarity (M) = moles / volume in L = 0.002 mol / 0.62 L = 0.00323 M
4. Since barium hydroxide dissociates into Ba2+ and 2 OH-, the concentration of OH- ions is twice the molarity of the barium hydroxide solution:
[OH-] = 2 * 0.00323 = 0.00646 M
5. Calculate the pOH:
pOH = -log([OH-]) = -log(0.00646) = 2.19
6. Calculate the pH:
pH + pOH = 14
pH = 14 - 2.19
pH = 11.81
Therefore, the pH of the solution containing 0.344 g of barium hydroxide in 620 mL of water is approximately 11.81.