a.) pH=-log[H+]=-log(0.10M HAc)
b.) At the 1/2 eq. pt., 1/2 of the HAc will react with NaOH. So, 25 z 10^-3L (0.01M)= moles of HAc
1/2*moles of HAc= moles of unreacted HAc
moles of unreacted HAc/total volume (37.5mL)= M of unreacted HAc
pH=-log[H+]=-log[M of unreacted HAc]
c.) Neutralization reaction, pH= 7
HAc + H20---> H30+ Ac
HAc= acetic acid, which is a weak acid.
Ka=1.8 x 10–5=[x][x]/0.100M-x
5% rule allows you to ignore -x
Ka=1.8 x 10–5=x^2/0.100M
solving for x,
x=1.34 x 10^-3
pH=-log(1.34 x 10^-3 M)
Concentrations of Ac and HAc are eqaul, pH=pKA.
Ac- + H2O ---> OH +HAc
I apologize about that one; I assumed HAc was a strong acid. Oppps.
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