I like to work in millimols. Let's just take 100 mL of each solution. To save space and typing let's call C2H5NH2 just RNH2.
millimols RNH2 = mL x M = 100 x 1.68 = 168
mmols HClO4 = 100 x 0.966 = 96.6
...RNH2 + HClO4 => RNH3^+ + ClO4^-
I..168.....0.........0.......0
add.......96.6................
C.-96.6.-96.6.......+96.6
E...71.4...0.........96.6
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH. Note that you need to convert Kb to pKa. Two steps to do that.
pKb = -log Kb, then
pKb + pKa = 14. Solve for pKa.
Calculate the pH of the buffer formed by mixing equal volumes [C_2H_5NH_2]= 1.68 M with [HClO_4]= 0.966 M . Kb=4.3×10−4.
1 answer
1 answer