Consider the titration of a 35 mL sample of 0.175 M HBr with 0.210 M KOH. Determ
ID: 1040958 • Letter: C
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