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    pH of solution and equilibrium

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    4). calculate the pH of a 0.265 M solution of propanoic acid
    HC3H5O2 (aq) = C3H5O2- (aq) + H+ (aq), Ka = 1.34 x 10^-5
    Propanoic acid ionizes in water according to the equation above
    A) write the equilibrium- constant expression for the reaction
    B) calculate the pH of a 0.265 M solution of propanoic acid

    5) A 0.496 g sample of sodium propanoate , NaC3H5O2 , is added to a 50.0ml sample of a 0.265 M solution of propanoic acid. Assuming that no change in the volume of the solution occurs , calculate each of the following
    (I) The concentration of propanoate ion. , in the solution
    (Ii) The concentration of the H+ (aq) ion I the solution

    6) The methanoate ion , HCO2- (aq) , reacts with water to form methanoic acid and OH- ion as shown in the following equation
    HCO2 - (aq). + H2O (l). = HCO2H (aq) + OH- (aq)
    given that OH- concentration is 4.18 x 10^ -6 M in a 0.309 M solution of sodium methanoate , calculate each of the following
    (I) The value of Kb for the methanoate ion
    (Ii). The value of Ka for the methanoic acid

    7) which acid is stronger, propanoic acid or methanoic acid? Justify your answer

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    Solution Summary

    It provides several examples of finding the pH of the solution.

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