A solution of barium hydroxide contains 5.2 x 10^25 hydroxide ions, how many grams of barium hydroxide were originally dissolved?
Barium hydroxide has a chemical formula of `Ba(OH)_2` . This means that each mole of barium hydroxide contains 1 mole of barium and 2 moles of hydroxide ions. Also, one mole of any substance contains an Avogadro's number of atoms (or sub-units), that is, 6.023 x 10^23 atoms.
Here, the solution contains 5.2 x 10^25 hydroxide ions.
Using unitary method, 6.023 x 10^23 atoms are in 1 mole, then 5.2 x 10^25 atoms would be in:
number of moles = `(5.2 xx 10^25)/(6.023 xx 10^23) = 86.34`
Thus, there are (about) 86.34 moles of hydroxide ions in the solution. This means, there are only (about) 43.2 moles (= 86.34/2) of barium in the solution and only 43.2 moles of barium hydroxide was dissolved initially.
Molecular weight of barium hydroxide is 171.34 gm/mole (= 137.34 + 2 x 16 + 2 x 1).
Thus the mass of barium hydroxide, initially dissolved = 43.2 x 171.34
= 7,401.9 grams.
Hence about 7,402 grams of barium hydroxide were dissolved originally.
Hope this helps.