The correct order of increasing [H_3O^{+}] in | vedclass

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(C) To determine the concentration of $[H_3O^{+}]$,we analyze the nature of the solutes in $0.01 \ M$ aqueous solutions:$1$. $NaNO_2$ is a salt of a w

Vedclass · Accepted Answer

$0.01 \ M \ NaNO_2 < 0.01 \ M \ NaCl < 0.01 \ M \ H_2S < 0.01 \ M \ H_2SO_4$

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(C) To determine the concentration of $[H_3O^{+}]$,we analyze the nature of the solutes in $0.01 \ M$ aqueous solutions:$1$. $NaNO_2$ is a salt of a weak acid $(HNO_2)$ and a strong base $(NaOH)$,making it basic ($pH > 7$,low $[H_3O^{+}]$).$2$. $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,making it neutral ($pH = 7$,$[H_3O^{+}] = 10^{-7} \ M$).$3$. $H_2S$ is a weak acid,which partially dissociates to produce $[H_3O^{+}]$.$4$. $H_2SO_4$ is a strong acid,which completely dissociates to produce a high concentration of $[H_3O^{+}]$.Thus,the increasing order of $[H_3O^{+}]$ is: $NaNO_2 < NaCl < H_2S < H_2SO_4$.

The correct order of increasing $[H_3O^{+}]$ in the following aqueous solutions is:

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