A.1. Write the phrase to establish equilibrium. A.2. Calculate the pH of the propanoic acid solution at a concentration of 0.265M. Show calculations B. Add 0.496 g of sodium propanoate, CH3CH2COONa , To 50 ml of solutionPropenoic acid at a concentration of 0.265M. Assuming the volume of the solution does not change, Calculate the pH of the resulting solution. Show calculations.
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A. Propanoic acid,
CH3CH2COOH, Reacts with water according to the given reaction:
??3??2???? + ?2? ⇄ ??3??2???− + ?3?+?? = 1.34 × 10−5
A.1. Write the phrase to establish equilibrium.
A.2. Calculate the pH of the propanoic acid solution at a concentration of 0.265M.
Show calculations
B. Add 0.496 g of sodium propanoate,
CH3CH2COONa , To 50 ml of solutionPropenoic acid at a concentration of 0.265M. Assuming the volume of the solution does not change, Calculate the pH of the resulting solution. Show calculations.
C. the Methanate ion HCOO- Reacts with water to a basic pH.
C.1. Write the reaction.
C.2. Calculate the Ka of the methanic acid, HCOOH, If in solution in concentration of 0.309M Of sodium methane solution, HCOONa, Are found 4.18*10-6 M OH- ions.
C.3. Who is the stronger acid, propane or methane? EXPLAIN.
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