To find the pH of a solution containing barium hydroxide, we first need to calculate the molarity of the barium hydroxide solution.
Step 1: Calculate the moles of barium hydroxide (Ba(OH)2)
Molar mass of Ba(OH)2 = 137.34 g/mol (Ba: 137.34 g/mol, O: 16.00 g/mol, H: 1.01 g/mol)
Number of moles = 0.344 g / 137.34 g/mol = 0.0025 mol
Step 2: Calculate the molarity of the barium hydroxide solution
Volume of solution = 620 mL = 0.62 L
Molarity = moles / volume
Molarity = 0.0025 mol / 0.62 L = 0.004 mol/L
Step 3: Calculate the pH of the solution
Barium hydroxide will dissociate into Ba2+ and OH- ions. Since Ba(OH)2 is a strong base, it will completely dissociate. Therefore, the concentration of OH- ions will be double the molarity of the solution.
OH- concentration = 2 * 0.004 mol/L = 0.008 mol/L
pOH = -log[OH-] = -log(0.008) = 2.09
Since pH + pOH = 14, pH = 14 - 2.09 = 11.91
Therefore, the pH of the solution containing 0.344 g of barium hydroxide in 620 mL of water is 11.91.
Find the pH of a solution containing 0.344 g of barium hydroxide in 620 mL of water.
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