Use the Henderson-Hasselbalch equation to obtain the initial pH. I have something like 9.19 but you need to confirm that and do it more accurately than I estimated.
Then work the addition part of the problem.
You started with 80 mL x 0.187 M NH4Cl = 13.12 mmoles.
You started with 80 mL x 0.164 = 14.96 mmoles. Now we add NaOH and I like to use an ICE chart.
........NH4^+ + OH^- ==> NH3 + H2O
initial 14.96....0......13.12......
change..-2.....+2mmoles..+2
final...12.96....0......15.12
Now substitute into the HH equation and solve for the new pH. I think 9.3 is close for the new pH.
what will be the pH change when 20 mL of .1M NaOH is added to 80 mL of a buffer solution consisting of .164M NH3 and .187M NH4Cl? Pka 9.25
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