Question

By how much will the pH change if 25.0 mL of 0.20 M NaOH is added to 0.500 L of the buffer that contains 0.25 M NH3 and 0.45 M NH4+ (Kb = 1.8 x 10-5)? The initial pH of the buffer is 9.00. Give your answer to two decimal places.

Answer 1

number of moles of NH3= 0.25Mx0.500L = 0.125 moles

number of moles of NH4+ = 0.45Mx0.500L=0.225 moles

NaOH= 25.0ml of 0.20M
number of moles of NaOH= 0.20Mx0.025L=0.005 moles
after addition of NaOH
number of moles of NH3 = 0.125+0.005 = 0.13moles
number of moles of NH4+= 0.225 - 0.005 =0.22 moles
Kb= 1.8x10^-5
-log(Kb) = -log(1.8x10^-5)
PKb = 4.74
POH= PKB + log(salt/base)
POH= 4.74 + log(0.22/0,13)
POH= 4.97
POH+PH=14
PH=14-POH
PH= 14-4.97
PH=9.03
PH= 9.00
Change in PH= 9.03-9.00=0.03
chnage in PH= 0.03.