A  B
100. mL of 0.200 M HCl is titrated with 0.250
M NaOH.
Part A
What is the pH of the solution after 50.0 mL of base has been added?
Express the pH numerically.

pH = 
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Part B
What is the pH of the solution at the equivalence point?
Express the pH numerically.

pH = 
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A)
This question involves strong acid and strong base
mol of NaOH added = M*V = 0.250 M * 50.0 mL = 12.5 mmol
mol of HCl added = M*V = 0.200 M * 100 mL = 20 mmol
12.5 mmol of each will react
remaining mol of HCl = 20.0 mmol  12.5 mmol = 7.5
mmol
final volume = 100 mL + 50 mL = 150 mL
[H+] = number of mol / total volume
= 7.5 mmol / 150 mL
= 0.05 M
use:
pH = log [H+]
= log (0.05)
= 1.30
Answer: 1.30
B)
At equivalence equal moles of acid and base have been added
pH will be 7
Answer: 7
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