25.00 mL of 0.250 M acetic acid (HC₂H30₂) is titrated with 25.00 mL of 0.250 M NaOH. The K₁ for acetic acid is 1.8 x 10-5. HC₂H3O2(aq) + OH(aq) → C₂H3O2 (aq) a. What is the pH before the addition of NaOH? b. What is the pH half way to the equivalence point? c. What is the pH at the equivalence point? + H₂O(1) A 0.2128 M solution of NaOH is used to titrate 37.00 mL of 0.1988 M HI. a.) Write a balanced net ionic equation for the reaction that takes place during the titration. b.) What are the spectator ions? c.) What is the pH of the solution before any NaOH is added? d.) What is the pH of the solution at the equivalence point?

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25.00 mL of 0.250 Macetic acid (HC₂H₂O₂) is titrated with 25.00 mL of 0.250 M NaOH.
The Ka for acetic acid is 1.8 x 10-5.
HC₂H3O2(aq) + OH(aq) → C₂H3O2 (aq) + H₂O(1)
a. What is the pH before the addition of NaOH?
b. What is the pH half way to the equivalence point?
c. What is the pH at the equivalence point?
A 0.2128 M solution of NaOH is used to titrate 37.00 mL of 0.1988 MHI.
a.) Write a balanced net ionic equation for the reaction that takes place during the titration.
b.) What are the spectator ions?
c.) What is the pH of the solution before any NaOH is added?
d.) What is the pH of the solution at the equivalence point?
Transcribed Image Text:25.00 mL of 0.250 Macetic acid (HC₂H₂O₂) is titrated with 25.00 mL of 0.250 M NaOH. The Ka for acetic acid is 1.8 x 10-5. HC₂H3O2(aq) + OH(aq) → C₂H3O2 (aq) + H₂O(1) a. What is the pH before the addition of NaOH? b. What is the pH half way to the equivalence point? c. What is the pH at the equivalence point? A 0.2128 M solution of NaOH is used to titrate 37.00 mL of 0.1988 MHI. a.) Write a balanced net ionic equation for the reaction that takes place during the titration. b.) What are the spectator ions? c.) What is the pH of the solution before any NaOH is added? d.) What is the pH of the solution at the equivalence point?
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