Assignment:
Q1. 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
Q2. 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
Q3. 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
Provide complete and step by step solution for the question and show calculations and use formulas.