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?
3 answers
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.
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.
9 answers