The reaction of the strong acid HCl with the weak base CH_3NH_2 is: HCl(aq) + CH

Question

The reaction of the strong acid HCl with the weak base CH_3NH_2 is: HCl(aq) + CH_3NH_2(aq) rightarrow CH_3NH_3^+ (aq) + Cl^- (aq) To compute the pH of the resulting solution if 15mL of 0.31 M acid is mixed with 58mL of 0.90M base we need to start with the stoichiometry. Let's do the stoich in steps: How many moles of acid? How many moles of base? What is the limiting reactant? How many moles of the excess reagent after reaction? What is the concentration of the excess reagent after reaction? What is the concentration of the pH active product after reaction?

Explanation / Answer

we know that

moles = molarity x volume (L)

so

moles of acid = 0.31 x 15 x 10-3

moles of acid (HCl) = 4.65 x 10-3

now

moles of base = 0.90 x 58 x 10-3

moles of base (CH3NH2) = 52.2 x 10-3

now

the reaction is

HCl + CH3NH2 ---> CH3NH3Cl

we can see that

moles of HCl required = moles of CH3NH2 = 52.2 x 10-3

but only 4.65 x 10-3 moles of HCl is present

so

HCl is the limiting reagent

and

CH3NH2 is the excess reactant

now

moles of CH3NH2 remaining = 52.2 x 10-3 - 4.65 x 10-3

moles of CH3Nh2 remaining = 47.55 x 10-3

so

moles of excess reactant after reaction is 47.55 x 10-3

now

final volume = 15 + 58 = 73 ml

now

Concentration = moles / volume (L)

so

conc of excess CH3NH2 = 47.55 x 10-3 / 73 x 10-3

conc of excess CH3NH2 = 0.65 M

now

the prodcut is CH3NH3Cl

now

moles of CH3NH3Cl formed = moles of HCl added = 4.65 x 10-3

conc of CH3NH3Cl = 4.65 x 10-3 / 73 x 10-3

conc of CH3NH3Cl = 0.0637 M

so

the concentration of pH active product is 0.0637 M

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