When an excess of PbI2 is placed in 1.0 L of 0.05 M KNO3 solution:
Calculate the pH of the titration solution after addition of 49.00 mL 0.1000 M HCl to 100.0 mL 0.0500 M solution of NaOH:
Concentration HCl = 0.1000 M
volume HCl = 49.0 mL
moles HCl added = (Concentration HCl) * (volume HCl)
moles HCl added = (0.1000 M) * (49.0 mL)
moles HCl added = 4.90 mmol
moles H+ = moles HCl added
moles H+ = 4.90 mmol
concentration NaOH = 0.0500 M
volume NaOH = 100.0 mL
moles NaOH added = (concentration NaOH) * (volume NaOH)
moles NaOH added = (0.0500 M) * (100.0 mL)
moles NaOH added = 5.00 mmol
moles OH- = moles NaOH added
moles OH- = 5.00 mmol
Since moles OH- > moles H+, therefore, solution will be basic
excess moles OH- = (moles OH-) - (moles H+)
excess moles OH- = (5.00 mmol) - (4.90 mmol)
excess moles OH- = 0.100 mmol
total volume = (volume HCl) + (volume NaOH)
total volume = (49.00 mL) + (100.0 mL)
total volume = 149.0 mL
[OH-] = (excess moles OH-) / (total volume)
[OH-] = (0.100 mmol) / (149.0 mmol)
[OH-] = 6.71 x 10-4 M
pOH = -log[OH-]
pOH = -log(6.71 x 10-4 M)
pOH = 3.17
pH = 14 - pOH
pH = 14 - 3.17
pH = 10.83
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