For a solution equimolar in HCN and NaCN, which statement is false? [H + ] is la

Question

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

[H+] is larger than it would be if only the HCN were in solution.

Addition of NaOH will increase [CN–] and decrease [HCN].

[H+] is equal to Ka.

This is an example of the common-ion effect.

Addition of more NaCN will shift the acid-dissociation equilibrium of HCN to the left.

[H+] is larger than it would be if only the HCN were in solution.

Addition of NaOH will increase [CN–] and decrease [HCN].

[H+] is equal to Ka.

This is an example of the common-ion effect.

Addition of more NaCN will shift the acid-dissociation equilibrium of HCN to the left.

Explanation / Answer

Answer is

[H+] is equal to Ka this statement is false

regarding common ion effect all other state ments are correct

Ka is the percentage of the dissociation of H+ which will depend on concentration of HCN whic was not equal to H+

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