partb a)A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate th
ID: 503231 • Letter: P
Question
partb a)A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH. Express your answer numerically. Part b) A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×105) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 15.0 mL of HNO3. Express your answer numerically.Part c) A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×105) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 33.0 mL of NaOH. Express your answer numerically.
Explanation / Answer
a.
HBr KOH
MA = 0.15M MB = 0.25M
VA = 50ml VB = 13ml
M = MAVA -MBVB/VA+VB
= 0.15*50-0.25*13/50+13 = 4.25/63 = 0.0674M
M = [H+]
PH = -log[H+]
= -log0.067 = 1.1739
B.
no of moles of NH3 = molarity * volume in L
= 0.2*0.075 = 0.015moles
no of moles of HNO3 = molarity * volume in L
= 0.5*0.015 = 0.0075 moles
NH3 + HNO3 --------------> NH4NO3
I 0.015 0.0075 0
C -0.0075 -0.0075 0.0075
E 0.0075 0 0.0075
POH = Pkb + log[NH4NO3]/[NH3]
= 4.75 + log0.0075/0.0075
= 4.75
PH = 14-POH
= 14-4.75 = 9.25
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