In one litre of 0.1 M acetic solution, one millimole of HCl is added. The pH of the solution is: (Ka(CH3COOH)=2×10−5) (log 2 = 0.3, log 3 = 0.5, log 5 = 0.7)
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Solution
CH3COOHKa=2×10−5⇌CH3COO−+H+
0.1M010−3M0.1M−xx(10−3+x)M Ka=2×10−5=x(10−3+x)0.1−x if we neglect x wrt to 10−3 then, x=2×10−3 So x is not negligible wrt to 10−3 but it is negligible wrt to 0.1 2×10−5=x(10−3+x)0.1⇒x=10−3M [H+]=10−3+10−3=2×10−3⇒pH=2.7