Chloroacetic acid, HC2H2O2Cl, is a stronger monoprotic acid than acetic acid. In a 0.10 M solution, the pH is 1.96.
Calculate the pKa for chloroacetic
acid.
pKa =
pH = -log [ H+ ]
-log[ H+ ] = 1.96
[ H+ ] = 1× 10^(-1.96 )
= 0.0109
HC2H2O2Cl <------> H+ + C2H2O2Cl
Ka = [ H+ ] [ C2H2O2Cl- ] / [ HC2H2OCl ]
[ H+ ] = [ C2H2O2Cl- ] = 0.0109M
[ HC2H2O2Cl ] = 0.1 - 0.0109M = 0.0891M
Therefore,
Ka = (0.0109)^2/0.0891= 1.33× 10^-3
pKa = -logKa
= - log(1.33×10^-3)
= 2.87
Chloroacetic acid, HC2H2O2Cl, is a stronger monoprotic acid than acetic acid. In a 0.10 M solution,...
Chloroacetic acid, HC2H2O2Cl, is a stronger monoprotic acid than acetic acid. In a 0.10 M solution, the pH is 1.96. Calculate the Kafor chloroacetic acid. Calculate the pKa for chloroacetic acid. Use the correct number of significant figures.
I can't figure this problem out. Please help! Chloroacetic acid, HC2H202CI, is a stronger monoprotic acid than acetic acid. In a 0.10 M solution, the pH is 1.96. Incorrect. Did you use the correct equilibrium expression? Did you use the correct number of significant figures? Calculate the Ka for chloroacetic acid 0.0013 Ka
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