50.050.0 mL solution of 0.1390.139 M KOHKOH is titrated with 0.2780.278 M HClHCl. Calculate the pH of the solution after the addition of each of the given amounts of HClHCl. 0.00 mLpH=0.00 mLpH=     6.00 mLpH=6.00 mLpH=     12.5 mLpH=12.5 mLpH=

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Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.91QP: A 50.0-mL sample of a 0.100 M solution of NaCN is titrated by 0.200 M HCl. Kb for CN is 2.0 105....
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A 50.050.0 mL solution of 0.1390.139 M KOHKOH is titrated with 0.2780.278 M HClHCl. Calculate the pH of the solution after the addition of each of the given amounts of HClHCl.
0.00 mLpH=0.00 mLpH=
 
 
6.00 mLpH=6.00 mLpH=
 
 
12.5 mLpH=12.5 mLpH=
 
 
20.0 mLpH=20.0 mLpH=
 
 
24.0 mLpH=24.0 mLpH=
 
 
25.0 mLpH=25.0 mLpH=
 
 
26.0 mLpH=26.0 mLpH=
 
 
29.0 mLpH=29.0 mLpH=
 
A 50.0 mL solution of 0.139 M KOH is titrated with 0.278 M HCI. Calculate the pH of the solution after the addition of each of
the given amounts of HCl.
0.00 mL
pH
6.00 mL
pH =
12.5 mL
pH =
20.0 mL
pH :
24.0 mL
pH
25.0 mL
pH
26.0 mL
pH =
29.0 mL
pH =
Transcribed Image Text:A 50.0 mL solution of 0.139 M KOH is titrated with 0.278 M HCI. Calculate the pH of the solution after the addition of each of the given amounts of HCl. 0.00 mL pH 6.00 mL pH = 12.5 mL pH = 20.0 mL pH : 24.0 mL pH 25.0 mL pH 26.0 mL pH = 29.0 mL pH =
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