HC2H3O2(aq) + H2O(l) ---------> H3O^+ (aq) + C2H3O2^-
(aq)
I
0.2
0
0
C
-x
+x
+x
E
0.2-x
+x
+x
Ka =
[H3O^+][C2H3O2^-]/[HC2H3O2]
1.8*10^-5 = x*x/0.2-x
1.8*10^-5*(0.2-x) = x^2
x = 0.0018
[H3O^+] = 0.0018M
PH = -log[H3O^+]
= -log0.0018
= 2.7447
th following reaction: K,-1.8 × 10-s at 25°C Submit Incorrect; Try Again; 5 attempts remaining Prade...
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