Question
A volume of 100mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence point.
5.00ml of 1.00M NaOH, 200 mL of 1.00M NaOH, 100 mL of 1.00M NaOH, 10,0 mL of 1.00 M NaOH, 150 mL of 1.00M NaOH, 50.0 mL of 1.00M NaOH
5.00ml of 1.00M NaOH, 200 mL of 1.00M NaOH, 100 mL of 1.00M NaOH, 10,0 mL of 1.00 M NaOH, 150 mL of 1.00M NaOH, 50.0 mL of 1.00M NaOH
Answers
HCl + NaOH ==> NaCl + H2O
mols HCl initally = M x L = 1.00 M x 0.1 L = 0.1 mol HCl.
NaOH
M x L = 1.00 x 0.005L = 0.005 mols must be before.
1.00 x 0.200L = 0.2 mol must be after (02 is more than 0.1 HCl).
1.00 x 0.100L =0.1 mol must be at the equivalence point.
Etc.
mols HCl initally = M x L = 1.00 M x 0.1 L = 0.1 mol HCl.
NaOH
M x L = 1.00 x 0.005L = 0.005 mols must be before.
1.00 x 0.200L = 0.2 mol must be after (02 is more than 0.1 HCl).
1.00 x 0.100L =0.1 mol must be at the equivalence point.
Etc.
Related Questions
A sample of 25.00 mL of vinegar is titrated with a standard 1.02 M NaOH solution. It was found that...
What is the molarity of the acetic acid if 0.5ml of a vinegar solution has been titrated with the 0....
A weak acid HA (pka=6.00) was titrated with 1.00M NaOh. The acid solution had a volume of 100ml an...
Ok.I have a titration experiment yesterday 0.2 M of KHP were used and the volume of the titra...