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
To classify the conditions based on whether they are before, at, or after the equivalence point, we need to understand what happens during a titration.
In a titration, a reaction between an acid and a base is carried out using a known concentration of one solution to determine the concentration of the other. The equivalence point is reached when stoichiometrically equivalent amounts of acid and base have reacted.
In this specific scenario, we are titrating 1.00 M HCl with 1.00 M NaOH. The balanced chemical equation for the reaction is:
HCl + NaOH → NaCl + H2O
Based on the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH, meaning that the ratio of HCl to NaOH is 1:1.
Now, let's analyze the quantities of NaOH solution added:
1. 5.00 mL of 1.00 M NaOH:
- Before the equivalence point. It is a small quantity compared to the initial volume of HCl (100 mL) and will not be enough to neutralize all the acid.
2. 200 mL of 1.00 M NaOH:
- After the equivalence point. The volume of NaOH added is much larger than the initial volume of HCl (100 mL), indicating that excess base is present.
3. 100 mL of 1.00 M NaOH:
- At the equivalence point. The volume of NaOH added is equal to the initial volume of HCl (100 mL), indicating that stoichiometrically equivalent amounts of acid and base have reacted.
4. 10.0 mL of 1.00 M NaOH:
- Before the equivalence point. Similar to the first case, it is a small quantity compared to the initial volume of HCl (100 mL).
5. 150 mL of 1.00 M NaOH:
- After the equivalence point. The volume of NaOH added is larger than the initial volume of HCl (100 mL), indicating an excess of base.
6. 50.0 mL of 1.00 M NaOH:
- Before the equivalence point. It is a small quantity compared to the initial volume of HCl (100 mL) and will not be enough to neutralize all the acid.
To summarize:
- Before the equivalence point: 5.00 mL of 1.00 M NaOH and 10.0 mL of 1.00 M NaOH.
- At the equivalence point: 100 mL of 1.00 M NaOH.
- After the equivalence point: 200 mL of 1.00 M NaOH, 150 mL of 1.00 M NaOH, and 50.0 mL of 1.00 M NaOH.
To determine whether each condition is before the equivalence point, at the equivalence point, or after the equivalence point, we need to compare the amounts of HCl and NaOH used.
1. 5.00 mL of 1.00 M NaOH: Before the equivalence point. At this point, less NaOH has been added than the stoichiometric amount required to neutralize all the HCl.
2. 200 mL of 1.00 M NaOH: After the equivalence point. At this point, more NaOH has been added than the stoichiometric amount required to neutralize all the HCl.
3. 100 mL of 1.00 M NaOH: At the equivalence point. At this point, the stoichiometric amount of NaOH required to neutralize all the HCl has been added.
4. 10.0 mL of 1.00 M NaOH: Before the equivalence point. Similar to the first condition, less NaOH has been added than the stoichiometric amount required to neutralize all the HCl.
5. 150 mL of 1.00 M NaOH: After the equivalence point. Similar to the second condition, more NaOH has been added than the stoichiometric amount required to neutralize all the HCl.
6. 50.0 mL of 1.00 M NaOH: Before the equivalence point. Similar to the first and fourth conditions, less NaOH has been added than the stoichiometric amount required to neutralize all the HCl.
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.