A 0.195 M weak acid solution has a pH of 2.94. What is the Ka?

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Answer 1 · 2018-04-11T00:02:25

#6.742 × 10^-6#

#"pH"# of an acid is given by

#"pH" = -log["H"^+]#

#2.94 = -log["H"^+]#

#["H"^+] = 10^-2.94\ "M" = 1.148 × 10^-3\ "M"#

# #

#["H"^+]# for a weak acid is given by

#["H"^+] = "c" α#

#α = (["H"^+]) / "c" = (1.148 × 10^-3\ cancel"M")/(0.195 cancel"M") = 5.88 × 10^-3#

# #

From Ostwald’s dilution Law

#"K"_"a" = "c" α^2 = 0.195 × (5.88 × 10^-3)^2 = 6.742 × 10^-6#