and0.0014mol C2H3O−20.050L=0.028MC2H3O−2.Finally we now need to setup an ICE chart,[HC2H3O2]£H+¤£C2H3O−2¤I0.05200.028C−xxxE0.052−0.028 +x.UsingKa=1.7×10−5=(x)(0.028 +x)(0.052−x)and solving the quadratic equation we getx=£H+¤=3.15×10−5.So,pH=−log¡3.15×10−5¢=4.50So a change in pH of only 0.05 units.We now calculate the pH for the situation of adding the HClto pure water.From above we see that 0.0001 mol of H+are produced.These moles are now in a total volume of 40 mL so,£H+¤=0.0001mol H+0.040mL.0025MHence a pH ofpH=
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This document was uploaded on 11/01/2011 for the course CHM 2046 at University of Florida.