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Author Question: A 100.0 mL sample of 0.20 mol L-1 HF is titrated with 0.10 mol L-1 KOH. Determine the pH of the ... (Read 51 times)

kfurse

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on: Nov 18, 2019

Question 1

An acid solution is 0.100 mol L-1 in HCl and 0.200 mol L-1 H2SO4. What volume of a 0.150 mol L-1 KOH solution would completely neutralize 500.0 mL of this solution?
◦ 1.67 L
◦ 1.02 L
◦ 1.98 L
◦ 0.250 L
◦ 0.700 L

Question 2

A 100.0 mL sample of 0.20 mol L-1 HF is titrated with 0.10 mol L-1 KOH. Determine the pH of the solution after the addition of 200.0 mL of KOH. The Ka of HF is 3.5 × 10-4.
◦ 7.00
◦ 3.46
◦ 9.62
◦ 8.14
◦ 10.54
Chemistry: A Molecular Approach (Canadian) ¦ Tro, Fridgen, Shaw ¦ 2nd Edition
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Marked as best answer by kfurse on Nov 18, 2019

Dinolord

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Reply #1 on: Nov 18, 2019
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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kfurse

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Reply 2 on: Nov 18, 2019
Wow, this really help

okolip

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

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