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.
PLS IF YOU CAN, TYPE THE ANSWER AND DO NOT WRITE IT DOWN, THANKS IN ADVANCE.
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.
Step by step
Solved in 2 steps