Mix together 5.0 mL of solution #1 – 1.0 M HA with 35 mL of solution #2 – 1.0 M NaA in a small Erlenmeyer flask. What are the concentrations of HA and A– in solution #7?
It appears from the problem that HA is a weak acid. If so, don't you need a Ka for HA? And is solution 7 the result of 1+2.
yes, solution 7 is the result of 1+2. So do I solve for Ka and use that equation?
If solution HA is a weak acid, I don't know how you solve for Ka unless it is given.
If soln HA is a strong acid, then
A from HA = 5 mL x 1 M = 5 millimoles diluted to 40 mL = 0.125 M
A from NaA is 35 mL x 1 M = 35 millimoles diluted to 40 mL = 0.875 M
Total A is 0.125 + 0.875 = 1 M (Technically it isn't permissible to add molarities but we can do it in this case. The correct way to do it is to add millimoles HA (5 x 1 = 5 millimoles) + millimoles NaA (35 mL x 1 M = 35 millimoles) and 5 + 35 = 40 millimoles in 40 mL = 40/40 = 1 M for A.
HA is zero if it is a strong acid.
To find the concentrations of HA and A- in solution #7, we need to use the principles of stoichiometry and the concept of dilution.
1. Determine the number of moles of HA in solution #1:
- Volume of solution #1 = 5.0 mL = 0.005 L
- Concentration of HA in solution #1 = 1.0 M
- Moles of HA in solution #1 = concentration * volume = 1.0 M * 0.005 L = 0.005 moles
2. Determine the number of moles of NaA in solution #2:
- Volume of solution #2 = 35 mL = 0.035 L
- Concentration of NaA in solution #2 = 1.0 M
- Moles of NaA in solution #2 = concentration * volume = 1.0 M * 0.035 L = 0.035 moles
3. As HA and NaA react together, they react in a 1:1 ratio to form A-. This means that each mole of HA will produce one mole of A-. So, the number of moles of A- formed is also 0.005 moles.
4. Calculate the total volume of solution #7:
- Volume of solution #1 = 5.0 mL = 0.005 L
- Volume of solution #2 = 35 mL = 0.035 L
- Total volume of solution #7 = Volume of solution #1 + Volume of solution #2 = 0.005 L + 0.035 L = 0.04 L
5. Now, we can calculate the concentration of HA and A- in solution #7:
- Concentration of HA in solution #7 = Moles of HA in solution #1 / Total volume of solution #7 = 0.005 moles / 0.04 L = 0.125 M
- Concentration of A- in solution #7 = Moles of A- formed / Total volume of solution #7 = 0.005 moles / 0.04 L = 0.125 M
Therefore, the concentrations of HA and A- in solution #7 are 0.125 M.