What are the concentrations of A– and HA after 24 mL of 0.33 M hydrochloric acid solution is added to 86 mL of a solution that has [A– ]=0.67 M? Do NOT consider any equilibrium reactions in this problem.
A–(aq) + HCl(aq) → HA(aq) + Cl–(aq)
How do I calculate this problem? I'm extremely lost in my class right now and have no idea how to do this.
To solve this problem, we need to use the concept of stoichiometry. Here's the step-by-step process:
Step 1: Write down the given information:
- Initial volume of HCl solution: 24 mL
- Initial volume of A- solution: 86 mL
- Concentration of A- solution: 0.67 M
- Concentration of HCl solution: 0.33 M
Step 2: Convert the volumes to liters:
- HCl solution = 24 mL = 0.024 L
- A- solution = 86 mL = 0.086 L
Step 3: Calculate the number of moles of A- and HCl:
- Moles of A- = Concentration * Volume = 0.67 M * 0.086 L
- Moles of HCl = Concentration * Volume = 0.33 M * 0.024 L
Step 4: Determine the limiting reactant:
- Compare the moles of A- and HCl from step 3. The reactant with fewer moles is the limiting reactant.
Step 5: Calculate the moles of the product:
- The balanced equation shows a 1:1 stoichiometric ratio between A- and HA. Therefore, if A- is the limiting reactant, the moles of HA formed will be equal to the moles of A-. If HCl is the limiting reactant, the moles of HA formed will be equal to the moles of HCl.
Step 6: Convert moles back to concentrations:
- Concentration of A- = Moles of A- / Total volume (A- solution + HCl solution)
- Concentration of HA = Moles of HA / Total volume (A- solution + HCl solution)
Now, let's calculate step by step, assuming A- is the limiting reactant:
Step 2: Convert the volumes to liters:
- HCl solution = 24 mL = 0.024 L
- A- solution = 86 mL = 0.086 L
Step 3: Calculate the number of moles of A- and HCl:
- Moles of A- = Concentration * Volume = 0.67 M * 0.086 L = 0.05762 moles
- Moles of HCl = Concentration * Volume = 0.33 M * 0.024 L = 0.00792 moles
Since the number of moles of HCl is less than the number of moles of A-, HCl is the limiting reactant in this case.
Step 5: Calculate the moles of the product:
- Moles of HA formed = Moles of HCl = 0.00792 moles
Step 6: Convert moles back to concentrations:
- Total volume = Volume of A- solution + Volume of HCl solution = 0.086 L + 0.024 L = 0.11 L
- Concentration of A- = Moles of A- / Total volume = 0.05762 moles / 0.11 L = 0.5246 M
- Concentration of HA = Moles of HA / Total volume = 0.00792 moles / 0.11 L = 0.072 M
Therefore, the concentrations of A- and HA after the reaction are:
- [A- ] = 0.5246 M
- [HA] = 0.072 M