A 0.100 m solution of chloroacetic acid (clch2cooh) is 11.0% ionized.

M 2 2 – M + M 2 M K a

Very first create the response:
ClCH2COOH <--> H+
+ClCH2COO-
could be written as:
Maybe Not
Ionized <--> Ionized H+
and ClCH2COO- tend to be ions ;0).
Which means you have got all the data you ought to determine Ka
andconcentrations of every:
[ClCH2COOH] = 0.1 M * [(100-11)/100] = 0.089
[H+] = 0.1 M * (11/100) = 1.1*10-2 M
[ClCH2COO-] = 0.1 M * (11/100)
=1.1*10-2 M
Ka = ([H+][ClCH2COO-])
/[ClCH2COOH]
∴Ka = (0.0112) / (0.89) = 1.36*10-3

The original focus of chloroacetic acid | ClCH.COOH|ー.İs 0. 100 M. percent ionization is 11.0percent Number of acid ionized is, 11.0 0.011 M 100 0.100 Mx 11.0percent = 0.100 M Set ICE dining table preliminary (M) Change(M): 0.011 Equilibrium(M): (0.100-0.011)00110011 0.100 +0.011+0.011 – 0.0890 CÍCH.COO- | = 0.01 1 M H’ | = 0.011 M [CICH,COOH-0.0890M 「CICH.c00][H]- [CICH,COOH! ). ( (0.0890 0.011 1.4×10-3 =
A 0.100 M solution of achloroacetic acid
(ClCH2COOH) is 11.0% ionized. Using this information,
calculateeach of the following.
[ClCH2COO-]M
[H+]M
[ClCH2COOH]MKa for chloroacetic acid

Leave a Comment