Asked by Jean
6. A titration was carried out to determine the concentration of 12.5 mL of an HCl(aq) solution. 9.0 mL of 0.050 mol/L barium hydroxide solution was required to reach the endpoint. Calculate the concentration of chloride ions in the reaction mixture at the end of the titration.
Answers
Answered by
DrBob222
2HCl + Ba(OH)2 ==> BaCl2 + 2H2O
For such small numbers I like to work in millimoles. That is a way to keep all those zeros.
mols Ba(OH)2 = M x L = 0.05 x 0.009 = 0.00045 but
millimols Ba(OH)2 = M x mL = 0.05 x 9 = 0.45
Convert mmols Ba(OH)2 to mmols HCl. From the equation 2 mol HCl = 1 mol Ba(OH)2; therefore, 2 x 0.045 = 0.9 millimols HCl
M HCl = M/L = mmols/mL = 0.9/12.5 = 0.072 M.
(HCl) = 0.072 so (Cl^-) must be 0.072 M also since there is 1 atom Cl per 1 molecule of HCl.
For such small numbers I like to work in millimoles. That is a way to keep all those zeros.
mols Ba(OH)2 = M x L = 0.05 x 0.009 = 0.00045 but
millimols Ba(OH)2 = M x mL = 0.05 x 9 = 0.45
Convert mmols Ba(OH)2 to mmols HCl. From the equation 2 mol HCl = 1 mol Ba(OH)2; therefore, 2 x 0.045 = 0.9 millimols HCl
M HCl = M/L = mmols/mL = 0.9/12.5 = 0.072 M.
(HCl) = 0.072 so (Cl^-) must be 0.072 M also since there is 1 atom Cl per 1 molecule of HCl.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.