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

Question

The reaction of the strong acid HBr with the strong base KOH is: HBr(aq) + KOH(aq) rightarrow HOH(l) + KBr(aq) Compute the pH of the resulting solution if 72mL of 0.79M acid is mixed with 22mL of 0.33M 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

Given that 72 mL of 0.79 M HBr and 22 mL of 0.33 M KOH.

moles = Volume in Litres x Molarity

a) Moles of acid before reaction

= 0.072 L x 0.79 M

= 0.057 mol

b) Moles of base before reaction

= 0.022 L x 0.33 M

= 0.0073 mol

c) Since moles of base are less than acid,

base is the limiting reagent.

d) how many moles of excess reagent after reaction

= 0.057 mol -  0.0073 mol

= 0.0497 mol

e) concentration of excess reagent after reaction

= moles/ total volume

= 0.0497 mol / ( 0.072 L + 0.022 L)

= 0.53 M

i.e. Since acid is in excess,

[H+] = 0.53 M

e)   

[H+] = 0.53 M

pH = -log [H+] = - log [0.53] = 0.27

Therefore,

pH of the resulting solution = 0.27

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