Question

A solution of 25.0 mL of 0.095 M hydroxylamonlum ion (weak acid) was titrated with 0.110 M NaOH. The Ka of hydroxylamonium is $1.1 \times 10^{-6}$. Calculate the following: What is the volume of NaOH solution at the equivalence point? 21.6 mL What is the pH after 10 mL of the NaOH is added? 10.85 What is the pH at the equivalence point? 9.85 What is the pH after 25 mL of the NaOH is added? 11.87

          A solution of 25.0 mL of 0.095 M hydroxylamonlum ion (weak acid) was titrated with 0.110 M NaOH. The Ka of hydroxylamonium is $1.1 \times 10^{-6}$. Calculate the following:
What is the volume of NaOH solution at the equivalence point? 21.6 mL
What is the pH after 10 mL of the NaOH is added? 10.85
What is the pH at the equivalence point? 9.85
What is the pH after 25 mL of the NaOH is added? 11.87

A solution of 25.0 mL of 0.095 M hydroxylamonlum ion (weak acid) was titrated with 0.110 M NaOH. The Ka of hydroxylamonium is 1.1 × 10^-6. Calculate the following:
What is the volume of NaOH solution at the equivalence point? 21.6 mL
What is the pH after 10 mL of the NaOH is added? 10.85
What is the pH at the equivalence point? 9.85
What is the pH after 25 mL of the NaOH is added? 11.87

Added by Todd L.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Shyam P

04/13/2022

Step 1

At the equivalence point, the moles of hydroxylammonium ion (weak acid) will be equal to the moles of NaOH added. Moles of hydroxylammonium ion = Molarity x Volume Moles of NaOH = Molarity x Volume Since the moles are equal at the equivalence point: 0.095 M x Show more…

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A solution of 25.0 mL of 0.095 M hydroxylammonium ion (weak acid) was titrated with 0.110 M NaOH. The Ka of hydroxylammonium is 1.1 x 10^(-6). Calculate the following: What is the volume of NaOH solution at the equivalence point: mL What is the pH after 10 mL of the NaOH is added? What is the pH at the equivalence point? What is the pH after 25 mL of the NaOH is added?