Answer to Question #78091 in General Chemistry for Ashley

Question

Consider the titration of a 35.0 mL sample of 0.180 molL?1 HBr with 0.210 molL?1 KOH. Determine each quantity: Part A the initial pH Express your answer using three decimal places. pH = Part B the volume of added base required to reach the equivalence point V = mL Part C the pH at 11.4 mL of added base Express your answer using three decimal places. pH = Part D the pH at the equivalence point Express your answer using two decimal places. pH = Part E the pH after adding 5.0 mL of base beyond the equivalence point Express your answer using two decimal places. pH =

How do I find the answer to this

Answer 1

pH = - log 0.180=0.745

moles HBr = 0.0350 L x 0.180 M=0.0063
moles KOH required = 0.0063
V = 0.0063/ 0.210 M=0.03 L = 30 mL
moles KOH = 0.0114 L x 0.210 M=0.0024
moles HBr in excess = 0.0063 - 0.0024=0.0039
total volume = 46.4 mL = 0.0464 L
[H+]= 0.0039 / 0.0464=0.0841 M
pH = 1.075
moles HBr = moles KOH : pH = 7.00
volume KOH added = 5 + 30 mL=35 mL = 0.035 L
moles KOH added = 0.035 L x 0.210 M=0.00735
moles OH- in excess = 0.00735 -0.0063 =0.00105
total volume = 0.07 L
[OH-]= 0.00105 / 0.07=0.015
pOH = 1.82
pOH = 12.18