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Author Question: A solution contains 0.250 M HA (Ka = 1.0 106) and 0.45 M NaA. What is the pH after 0.29 mole of HCl ... (Read 32 times)

dejastew

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on: Mar 21, 2021

Question 1

A solution containing 10. mmol of and 5.0 mmol of is titrated with 1.5 M HCl. What total volume of HCl must be added to reach the second equivalence point?

◦ 10.0 mL
◦ 6.7 mL
◦ 3.3 mL
◦ 16.7 mL
◦ 21.7 mL

Question 2

A solution contains 0.250 M HA (Ka = 1.0 10–6) and 0.45 M NaA. What is the pH after 0.29 mole of HCl is added to 1.00 L of this solution?

◦ 0.54
◦ 7.47
◦ 5.47
◦ 2.09
◦ 8.53
Chemistry: An Atoms First Approach ¦ Zumdahl, DeCoste ¦ 3rd Edition
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Marked as best answer by dejastew on Mar 21, 2021

Madisongo23

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Reply #1 on: Mar 21, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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dejastew

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Reply 2 on: Mar 21, 2021
Wow, this really help

hramirez205

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Reply 3 on: Yesterday
Excellent

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