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anonymous

  • one year ago

Enough of a monoprotic acid is dissolved in water to produce a 0.0116 M solution. The pH of the resulting solution is 2.35. Calculate the Ka for the acid.

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  1. RamiroCruzo
    • one year ago
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    Convert pH 2.68 to (H^+) with pH = -log(H^+). Then HA --> H^+ + A^- Ka = (H^+)(A^-)/(HA) Substitute for (H^+) and (A^-). For (HA) substitute 0.0169-(H^+)

  2. Rushwr
    • one year ago
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    monoprotic acid is an acid that can donate only one proton. So this acid is like HCl. \[HCl \rightarrow H ^{+} + Cl ^{-}\] We know the concentration of HCl solution. \[K _{a} = \frac{ [H ^{+}][Cl ^{-}] }{ [HCl] }\] \[[H ^{+}] = [Cl ^{-}]\] We can find H^+ concentration using pH \[pH = -\log_{10} [H ^{+}]\] HCl concentration is 0.0116M So \[K _{a}= \frac{ [H ^{+}]^{2} }{ [HCl] }\]

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