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Barium hydroxide forms several hydrates. A specimen of barium hydroxide, suspected of being a hydrate, was prepared and analyzed as follows to determine its formula.
3.632 g of the compound was dissolved in water to give 250.0mL of solution. 25.00 mL of this solution was titrated with 0.0987 M of HCl, using methyl-orange as indicator. Precise titrations of 23.34, 23.26 and 23.29 mL of HCl were obtained. Determine the formula of the hydrate.
So far I got...
(23.34, 23.26 and 23.29)/3 = 23.30mL
Ba(OH)2 + 2HCl --> BaCl2 + 2H2O
3.632g/0.250L * (1/171.34g) = 0.08479 mols/L Ba(OH)2
0.08479*0.025L = 0.00212 mols Ba(OH)2
0.0987 * 0.0233L = 0.00230mols HCl
Ans now I have no clue what to do...
3.632 g of the compound was dissolved in water to give 250.0mL of solution. 25.00 mL of this solution was titrated with 0.0987 M of HCl, using methyl-orange as indicator. Precise titrations of 23.34, 23.26 and 23.29 mL of HCl were obtained. Determine the formula of the hydrate.
So far I got...
(23.34, 23.26 and 23.29)/3 = 23.30mL
Ba(OH)2 + 2HCl --> BaCl2 + 2H2O
3.632g/0.250L * (1/171.34g) = 0.08479 mols/L Ba(OH)2
0.08479*0.025L = 0.00212 mols Ba(OH)2
0.0987 * 0.0233L = 0.00230mols HCl
Ans now I have no clue what to do...