The idea here is the Ba(OH)2 is neutralized by the HCl BUT more acid is added than is necessary. So you must determine how much of an excess HCl was added. You do that by titrating with NaOH. However, the strength of the NaOH is not given to you. In the separate titration of NaOH you determine the molarity (mols/dm3) of the NaOH.
In the separate titration with NaOH, we have 28.5 cc of the 0.200 M HCl solution reacting with 25.0 cc of the NaOH. Determine the mols/dm3 = M) of the NaOH.
...................NaOH + HCl ==> NaCl + H2O
millimols HCl = mL x M = 28.5 x 0.2 = 5.7
millimols NaOH must be 5.7 since the equation shows 1 mol HCl = 1 mol NaOH.
Then M NaOH = millimols/mL = 5.7/25.0 = 0.228 M.
.................Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
millimols HCl added initially = mL x M = 100 mL x 0.200 M = 20.0
Excess HCl added = millimols NaOH added = 10.9 mL x 0.228 M = 2.485 millimols HCl excess. So how much Ba(OH)2 was present.
Total HCl added = 20.0 millimols.
- excess HCl ......= 2.485
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20.0 - 2.485 = 17.5 millimols HCl.
Millimols Ba(OH)2 is 17.5 x (1 mol Ba(OH)2/2 mol HCl) = 8.76 mmols = 0.00876mols.
Convert to grams Ba(OH)2. g Ba(OH)2 = mols Ba(OH)2 x M Ba(OH)2 = ?
Then % Ba(OH)2 = (grams Ba(OH)2/1.6524)*100 = ?
Check my work, especially the arithmetic. Post your work if you get stuck.
An impure sample of barium hydroxide of mass 1.6524 g was allowed to react with 100 cm3 of 0.200 mol dm-3hydrochloric acid. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of sodium hydroxide solution was required. 25.0 cm3 of the sodium hydroxide required 28.5 cm3 of the hydrochloric acid in a separate titration. Calculate the percentage purity of the sample of barium hydroxide.
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