The reaction of the strong acid HBr with the strong base KOH is: HBr(aq) + KOH(l

Question

The reaction of the strong acid HBr with the strong base KOH is: HBr(aq) + KOH(l) rightarrow HOH(l) + KBr [aq). Compute the pH of the resulting solution if 72mL of 0.70M acid is mixed with 29mL of 0.35M base. Let's do this in steps: How many moles of acid before reaction? How many moles of base before reaction? 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 pH of the resulting solution?

Explanation / Answer

1. Acid: For HBr, Volume = 72 ml = 0.072 L

Concentration = 0.70 M

Moles of acid = concentration*volume = 0.70*0.072 = 0.0504 mol

moles of acid = 0.0504 mol

2. Base: For KOH, Volume = 29 ml = 0.029 L

Concentration = 0.35 M

Moles of base = concentration*volume = 0.35*0.029 = 0.0101 mol

moles of base = 0.0101 mol

3. Limiting reagent : KOH

KOH has lesser moles, so it is limiting reagent

4. Moles of excess reagent after reaction: moles of acid-moles of base = 0.0504-0.0101 = 0.0403 mol

5. Moles of excess reagent = moles of acid left = 0.0403 mol

Total volume = 0.072+0.029 = 0.101 L

concentration = moles/volume = 0.0403/0.101 = 0.40 M

6. HBr is left, and it is strong acid, so whole of acid will dissociate.

[HBr] = [H+] = 0.40 M

pH = -log[H+] = -log(0.40) = 0.40

pH = 0.40

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