a 20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution.
ID: 501704 • Letter: A
Question
a 20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the PH of the solution after the following volumes of base have been added.A. 14.0 mL B. 19.8 mL C. 20.0 mL D. 34.0 ML a 20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the PH of the solution after the following volumes of base have been added.
A. 14.0 mL B. 19.8 mL C. 20.0 mL D. 34.0 ML a 20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the PH of the solution after the following volumes of base have been added.
A. 14.0 mL B. 19.8 mL C. 20.0 mL D. 34.0 ML
Explanation / Answer
SOLUTION:
A.
HBr and NaOH dissociate in water to produce H+ and OH- respectively.
Moles of HBr = Molarity X Volume = 0.2 X 20mL = 4 mmol
Moles of NaOH added = 0.2mL X 14mL = 2.8 mmol
Moles of H+ left unreacted = 4 - 2.8 = 1.2 mmol = 0.0012 moles ( because 1mmol = 1/1000 = 0.001 mole)
Total volume = 20mL + 14mL = 34 mL = 0.034 L ( because 1mL = 1/ 1000 = 0.001L)
Molarity of H+ = number moles / volume in liters = 0.0012 / 0.034L = 0.035 M
pH = -log[H+] = -log[0.035] = 1.45
B. Moles of NaOH added = 19.8mL X 0.2 = 3.96mmoles
moles of H+ unreacted = 4.0mmol - 3.96mmol = 0.04 mmol = 0.00004 moles
Total volume = 20mL + 19.8mL = 39.8mL = 0.0398L
Molarity of H+ = 0.00004 moles / 0.0398L = 0.001
pH = -log(0.001) = 3.0
C. Moles of NaOH added = 0.2 M X 20mL = 4mmol
All the H+ will be consumed as a result the solution will be neutral having pH = 7
D. Moles of NaOH added = 0.2 X 34.0mL = 6.8 mmol
Out of 6.8 mmol of OH- 4 mmol will react with 4 mmol of H+ leaving 2.8 mmol (0.0028 moles) of OH- unreacted
Total volume = 20mL + 34mL = 54 mL = 0.054L
Molarity of OH- = 0.0028 / 0.54L = 0.005M
pOH = -log[OH-] = -log(0.005) = 2.3
pH + pOH = 14
pH = 14 - 2.3 = 11.7
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