The pH at the beginning is
pH = pKa + log(base)/(acid)
Substitute pKa, base, and acid, and solve for initial pH.
You have 100 mL 0.100M NH3 and 0.100M NH4Cl = 10 millimoles of each. You're adding 5.00 mL of 0.100M HCl = 0.500 millimols.
.......NH3 + H^+ ==> NH4^+ + H2O
I......10.....0.......10........
add...........0.5...............
C......-0.5...-0.5.....+0.5
E......9.5.....0.......10.5
Now substitute the E line into the HH equation and solve for pH, then take the difference between the two pH values to arrive at the difference.
Calculate the change in pH when 5.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).
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