The Ka for HOBr is 2.8 x 10 ^-9. You titrate 20.0 mL of a 0.400 M HOBr solution using a 0.400 M NaOH. What is the pH after 10.0 mL of NaOH have been added (1/2 equivalence pt) and after 22.0 mL of NaOH have been added (overshot endpoint)?
pH after 10.0 mL of 0.400 M NaOH =
pH after 22.0 mL of 0.400 M NaOH = (total volume = 42.0 mL)
1)when 10.0 mL of NaOH is added
Given:
M(HOBr) = 0.4 M
V(HOBr) = 20 mL
M(NaOH) = 0.4 M
V(NaOH) = 10 mL
mol(HOBr) = M(HOBr) * V(HOBr)
mol(HOBr) = 0.4 M * 20 mL = 8 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.4 M * 10 mL = 4 mmol
We have:
mol(HOBr) = 8 mmol
mol(NaOH) = 4 mmol
4 mmol of both will react
excess HOBr remaining = 4 mmol
Volume of Solution = 20 + 10 = 30 mL
[HOBr] = 4 mmol/30 mL = 0.1333M
[OBr-] = 4/30 = 0.1333M
They form acidic buffer
acid is HOBr
conjugate base is OBr-
Ka = 2.8*10^-9
pKa = - log (Ka)
= - log(2.8*10^-9)
= 8.553
use:
pH = pKa + log {[conjugate base]/[acid]}
= 8.553+ log {0.1333/0.1333}
= 8.553
Answer: 8.55
2)when 22.0 mL of NaOH is added
Given:
M(HOBr) = 0.4 M
V(HOBr) = 20 mL
M(NaOH) = 0.4 M
V(NaOH) = 22 mL
mol(HOBr) = M(HOBr) * V(HOBr)
mol(HOBr) = 0.4 M * 20 mL = 8 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.4 M * 22 mL = 8.8 mmol
We have:
mol(HOBr) = 8 mmol
mol(NaOH) = 8.8 mmol
8 mmol of both will react
excess NaOH remaining = 0.8 mmol
Volume of Solution = 20 + 22 = 42 mL
[OH-] = 0.8 mmol/42 mL = 0.019 M
use:
pOH = -log [OH-]
= -log (1.905*10^-2)
= 1.7202
use:
PH = 14 - pOH
= 14 - 1.7202
= 12.2798
Answer: 12.28
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