What is the ph of a solution prepared by mixing 50.00 ml of 0.10 m methylamine, ch3nh2, with 20.00 ml of 0.10 m methylammonium chloride, ch3nh3cl? assume that the volume of the solutions are additive and that kb = 3.70 × 10-4 for methylamine

pH=10.97

Explanation:

the solution of methyl amine with methylammonium chloride will make a buffer solution.

The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:

pH = 14- pOH

Let us calculate pOH

[Salt] = [methylammonium chloride] = 0.10 M (initial)

After adding base

[base] = [Methylamine]=0.10

After mixing with salt

pKb= -log[Kb]= 3.43

Putting values

pOH =

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

there all right good

explanation: