If a solution of barium hydroxide contains 9.1x1025 hydroxide ions, how many grams of barium hydroxide were originally dissolved?
To solve this problem, we need to use the concept of molar mass and Avogadro's number.
First, let's determine the molar mass of barium hydroxide (Ba(OH)2).
Barium (Ba) has a molar mass of 137.33 g/mol.
Oxygen (O) has a molar mass of 16.00 g/mol.
Hydrogen (H) has a molar mass of 1.01 g/mol.
Since there are two hydroxide ions (OH-) in barium hydroxide, we multiply the molar mass of OH- by 2.
Oxygen (O) has a molar mass of 16.00 g/mol, and hydrogen (H) has a molar mass of 1.01 g/mol, giving us a total of 17.01 g/mol for OH-.
Next, we need to use Avogadro's number to convert the given number of hydroxide ions to moles.
Avogadro's number is approximately 6.022 x 10^23 ions/mole.
Dividing the given number of hydroxide ions (9.1 x 10^25) by Avogadro's number gives us:
9.1 x 10^25 ions / (6.022 x 10^23 ions/mole) = 151.41 moles of OH-
Finally, we can calculate the mass of the barium hydroxide by multiplying the number of moles by the molar mass:
Mass = Number of moles × Molar mass
Mass = 151.41 moles × (137.33 g/mol + 2 × 17.01 g/mol)
Mass ≈ 151.41 moles × 171.35 g/mol
Mass ≈ 25939.86 g
Therefore, the original amount of barium hydroxide dissolved was approximately 25939.86 grams.