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Consider the titration of a solution of 20.00 mL of 0.450 M NaOH with 0.250 M HCl.

(a) What is the pH of the solution after 20.00 mL of 0.250 M HCl were added?
a) pH = 1.78
b) pH = 2.72
c) pH = 13.22
d) pH = 7.00

(b) How many mL of 0.250 M HCl are required to reach the equivalence point?
a) 40.00 mL
b) 20.00 mL
c) 80.00 mL
d) 60.00 mL

(c) What is the pH of the solution after the addition of 40.00 mL of 0.250 M HCl?
a) pH = 1.78
b) pH = 2.72
c) pH = 13.22
d) pH = 7.00

1 Answer

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Final answer:

The titration question asks to calculate pH values at different stages of adding a strong acid to a strong base, and to determine the volume of acid needed to reach the equivalence point. The pH can be found by considering the moles of acid and base and their respective concentrations. The equivalence point is found where moles of acid equal moles of base.

Step-by-step explanation:

The question involves a titration of a strong acid (HCl) with a strong base (NaOH). The reaction that occurs during the titration is:
HCl + NaOH → NaCl + H₂O

The moles of NaOH initially present in the solution can be calculated as follows:
Moles(NaOH) = 20.00 mL × 0.450 M = 0.009 moles

The moles of HCl that have been added after 20.00 mL is:
Moles(HCl) = 20.00 mL × 0.250 M = 0.005 moles

(a) Since the moles of HCl added are less than the moles of NaOH initially present, the reaction has not yet reached the equivalence point. The remaining NaOH will determine the pH of the solution, which can be calculated by finding the molarity of the unreacted NaOH and then determining the pOH, followed by calculating the pH.

(b) To find the volume of 0.250 M HCl required to reach the equivalence point, you would use the initial moles of NaOH and set them equal to the moles of HCl that need to be added. Since both reagents react in a 1:1 molar ratio:
0.009 moles NaOH = Moles HCl
Volume HCl required = 0.009 moles / 0.250 M

(c) At the equivalence point, the moles of HCl added would completely neutralize the moles of NaOH. Beyond this point, any additional HCl would make the solution acidic, with the pH corresponding closely to the pH of the HCl solution being added.

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