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Author Question: For a solution equimolar in HCN and NaCN, which statement is false? (Read 37 times)

geodog55

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For a solution equimolar in HCN and NaCN, which statement is false?

This is an example of the common ion effect.
The [H+] is larger than it would be if only the HCN was in solution.
The [H+] is equal to the Ka.
Addition of more NaCN will shift the acid dissociation equilibrium of HCN to the left.
Addition of NaOH will increase [CN] and decrease [HCN].


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Marked as best answer by geodog55 on Mar 21, 2021

AmberC1996

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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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geodog55

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Reply 2 on: Mar 21, 2021
:D TYSM


AmberC1996

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Reply 3 on: Yesterday
Gracias!

 

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