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Science Clinic => Chemistry => Topic started by: karlynnae on Mar 21, 2021

Title: Consider the titration of 100.0 mL of 0.10 M H2A (Ka1 = 1.50 104; Ka2 = 7.82 107) with 0.20 M ...
Post by: karlynnae on Mar 21, 2021

Question 1

Consider the titration of 100.0 mL of 0.10 M H₂A (K_a1 = 1.5 × 10^–4; K_a2 = 8.0 × 10^–7) with 0.20 M NaOH. Calculate the volume of 0.20 M NaOH required to reach an [H⁺] of 6.0 × 10^–4M.

◦ 0 mL
◦ 10. mL
◦ 25. mL
◦ 50. mL
◦ 65. mL

Question 2

Consider the titration of 100.0 mL of 0.10 M H₂A (K_a1 = 1.50 × 10^–4; K_a2 = 7.82 × 10^–7) with 0.20 M NaOH. Calculate the [H⁺] after 75.0 mL of 0.20 M NaOH has been added.

◦ 1.6 × 10^–6M
◦ 1.3 × 10^–8M
◦ 3.9 × 10^–7M
◦ 7.8 × 10^–7M
◦ None of these are correct.

Title: Consider the titration of 100.0 mL of 0.10 M H2A (Ka1 = 1.50 104; Ka2 = 7.82 107) with 0.20 M ...
Post by: Alyson.hiatt@yahoo.com on Mar 21, 2021

Answer 1

10. mL

Answer 2

7.8 × 10^–7M

Title: Re: Consider the titration of 100.0 mL of 0.10 M H2A (Ka1 = 1.50 104; Ka2 = 7.82 107) with 0.20 M
Post by: Ruohan Liu on Apr 28, 2024

Thank you！！