Item 11 Consider the titration of a 38.0 mL sample of 0.180 molL1 HBr with 0.200
ID: 1057782 • Letter: I
Question
Item 11
Consider the titration of a 38.0 mL sample of 0.180 molL1 HBr with 0.200 molL1 KOH. Determine each quantity:
Part A
the initial pH
Express your answer using three decimal places.
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Part B
the volume of added base required to reach the equivalence point
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Part C
the pH at 10.4 mL of added base
Express your answer using three decimal places.
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Part D
the pH at the equivalence point
Express your answer using two decimal places.
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Part E
the pH after adding 5.0 mL of base beyond the equivalence point
Express your answer using two decimal places.
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pH =Explanation / Answer
part-A
HBr -------> H+ +Br-
0.18M 0.18M
PH = -log[H+]
= -log0.18
= 0.7447
no of moles of HBr = molarity * volume in L
= 0.18*0.038 = 0.00684moles
part-B
no of moles of KOH = 0.00684 moles
volume of KOH = no of moles/molarity
= 0.00684/0.2 = 0.0342L = 34.2ml
part-c no of moles of KOH at 10.4 ml
no of moles of KOH = 0.2*0.0104 = 0.00208 moles
excess no of moles of HBr = 0.00684-0.00208 = 0.00476 moles
molarity of HBr = no ofmoles/total volume in L
= 0.00476/0.0484 = 0.09834M = [H+]
PH = -log[H+]
= -log0.09834 = 1.0072
PH = 7 at the equivalent point
part-D
no of moles of OH- = molarity * volume in L
= 0.2 *0.005 = 0.001moles
volume of KOH = 5 + 34.2 = 39.2 ml = 0.0392L
[OH-] = 0.001/0.0392 = 0.0255M
POH = -log0.0255
= 1.5934
PH = 14-POH
= 14-1.5934 = 12.4066
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