Asked by Mandy
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).
Answers
Answered by
DrBob222
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.
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.
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