Consider the titration of a 35 mL sample of 0.175 M HBr with 0.210 M KOH. Determ

Question

Consider the titration of a 35 mL sample of 0.175 M HBr with 0.210 M KOH. Determine :
1) the initial pH 2) the volume of added base required to reach the equivalence point 3) the pH of 12 mL of added base 4) the pH at equivalence point 5) the pH after adding 5 ml of base beyond the equivalence point Consider the titration of a 35 mL sample of 0.175 M HBr with 0.210 M KOH. Determine :
1) the initial pH 2) the volume of added base required to reach the equivalence point 3) the pH of 12 mL of added base 4) the pH at equivalence point 5) the pH after adding 5 ml of base beyond the equivalence point
1) the initial pH 2) the volume of added base required to reach the equivalence point 3) the pH of 12 mL of added base 4) the pH at equivalence point 5) the pH after adding 5 ml of base beyond the equivalence point

Explanation / Answer

1)

The reaction happening will be

HBr + KOH ----- KBr + H2O

pH = -log(H+) = -log(0.175) = 0.7569

2)

At equivalence point

number of moles of acid = number of moles of base

35/1000 * 0.175 = V/1000 * 0.210

V = 29.17 mL

3)

Number of moles of acid = 35/1000 * 0.175 = 0.006125 moles

Number of moles of base = 12/1000 * 0.210 = 0.00252 moles

Number of moles of acid left = 0.006125 - 0.00252 = 0.003605 moles

Volume of solution = 35 + 12 = 47 mL

Molarity of H+ = 0.003605/47 * 1000 = 0.0767

pH = -log(0.0767) = 1.115

4)

pH at equivalence point will be 7, since the titration of strong acid and strong base

5)

Number of moles of acid = 35/1000 * 0.175 = 0.006125 moles

Number of moles of base = 34.17/1000 * 0.210 = 0.0071757 moles

Number of moles of base left = 0.0071757 - 0.006125 = 0.0010507 moles

Volume of solution = 35 + 34.17 = 69.17 mL

Molarity of OH- = 0.0010507/69.17 * 1000 = 0.0151901113

pOH = -log(0.0151901113) = 1.818

pH = 14 - 1.818 = 12.185

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