In: Chemistry
Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid: (a) 5.00 mL (b) 29.50 mL (c) 36.00 mL
a)
Given:
M(HBr) = 0.1 M
V(HBr) = 5 mL
M(KOH) = 0.1 M
V(KOH) = 30 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.1 M * 5 mL = 0.5 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 30 mL = 3 mmol
We have:
mol(HBr) = 0.5 mmol
mol(KOH) = 3 mmol
0.5 mmol of both will react
remaining mol of KOH = 2.5 mmol
Total volume = 35.0 mL
[OH-]= mol of base remaining / volume
[OH-] = 2.5 mmol/35.0 mL
= 0.0714 M
use:
pOH = -log [OH-]
= -log (7.143*10^-2)
= 1.1461
use:
PH = 14 - pOH
= 14 - 1.1461
= 12.8539
b)
Given:
M(HBr) = 0.1 M
V(HBr) = 29.5 mL
M(KOH) = 0.1 M
V(KOH) = 30 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.1 M * 29.5 mL = 2.95 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 30 mL = 3 mmol
We have:
mol(HBr) = 2.95 mmol
mol(KOH) = 3 mmol
2.95 mmol of both will react
remaining mol of KOH = 0.05 mmol
Total volume = 59.5 mL
[OH-]= mol of base remaining / volume
[OH-] = 0.05 mmol/59.5 mL
= 0.0008 M
use:
pOH = -log [OH-]
= -log (8.403*10^-4)
= 3.0755
use:
PH = 14 - pOH
= 14 - 3.0755
= 10.9245
c)
Given:
M(HBr) = 0.1 M
V(HBr) = 36 mL
M(KOH) = 0.1 M
V(KOH) = 30 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.1 M * 36 mL = 3.6 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 30 mL = 3 mmol
We have:
mol(HBr) = 3.6 mmol
mol(KOH) = 3 mmol
3 mmol of both will react
remaining mol of HBr = 0.6 mmol
Total volume = 66.0 mL
[H+]= mol of acid remaining / volume
[H+] = 0.6 mmol/66.0 mL
= 0.0091 M
use:
pH = -log [H+]
= -log (9.091*10^-3)
= 2.0414