hmaritza149
hmaritza149 hmaritza149
  • 18-03-2019
  • Chemistry
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A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 16.0 mL of KOH.

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yousifmona44
yousifmona44 yousifmona44
  • 18-03-2019

moles HBr = 0.0500 L x 0.15 M=0.0075 

moles KOH = 0.0160 L x 0.25 M=0.0040 

moles H+ in excess = 0.0075 - 0.0040= 0.0035 

total volume = 0.070 L 

[H+]= 0.0035/0.070=0.050 M 

pH =1.55

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