First use the Henderson-Hasselbalch equation and solve for pKa.
pH = pKa + log (base)/(acid)
10.30 = pKa + log [0.4/0.2]
pKa = ?
mmols B initially = 0.4 x 250 = 100
mmols BH^+ initially = 0.2 x 250 = 50
..........B + H^+ ==> BH^+
I........100...0......50
add............25..........
C........-25...-25....+25
E.........75...0.......75
Using pKa from above, substitute the E line into a new HH equation and solve for pH.
Since B = BH^+ then the log term will be zero and the new pH will equal pKa.
A solution containing 0.40 M B (a weak base) and 0.20 M BH+ (the salt of the weak
base) has a pH = 10.30. What is the pH after 25 mmol of HCl is added to 250 mL of this
solution?
2 answers
10