Question

the reaction of the strong acid HBr with the strong base KOH is:

HCl(aq) + CH₃NH₂(aq) --> CH₃NH⁺₂ (aq) + Cl^-(aq)

To compute the pH of the resulting solution if 39mL of 0.45M acid is mixed with 51mL of 0.76M base. Let's do the stoich 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 are the moles of the pH active product after reaction?

Answer 1

Given:

M(HBr) = 0.45 M

V(HBr) = 39 mL

M(KOH) = 0.76 M

V(KOH) = 51 mL

1)

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.45 M * 39 mL = 17.55 mmol

Answer: moles of acid = 0.01755 mol

2)

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.76 M * 51 mL = 38.76 mmol

Answer: moles of acid = 0.03876 mol

3)

We have:

mol(HBr) = 17.55 mmol

mol(KOH) = 38.76 mmol

17.55 mmol of both will react

Answer: HBr is limiting

4)

remaining mol of KOH = 21.21 mmol

Answer: 0.02121 mol of KOH remains

5)

Total volume = 90.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 21.21 mmol/90.0 mL

= 0.236 M

Answer: concentration = 0.236 M

6)

use:

pOH = -log [OH-]

= -log (0.236)

= 0.628

use:

PH = 14 - pOH

= 14 - 0.628

= 13.4

Answer: 13.4