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Author Question: A 0.0925 g sample of a monoprotic base (mm = 17.03 g/mol) was dissolved in water to produce 100 mL ... (Read 44 times)

elizabeth18

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Question 1

0.653 g of a monoprotic acid (MW= 157 g/mol) is dissolved in water to produce 50.0 mL of a solution with pH = 2.13. Determine the ionization constant of the acid.
◦ 7.9 × 10-3
◦ 8.9 × 10-2
◦ 6.6 × 10-4
◦ 3.9 × 10-2
◦ 3.6 × 10-6

Question 2

A 0.0925 g sample of a monoprotic base (mm = 17.03 g/mol) was dissolved in water to produce 100 mL of solution with a pH = 11.00.  What is the ionization constant of this base?
◦ 1.8 × 10-6
◦ 1.8 × 10-11
◦ 1.8 × 10-21
◦ 1.8 × 10-5
◦ 1.1 × 10-6


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Marked as best answer by elizabeth18 on Jul 8, 2021

1_Step_At_ATime

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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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