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N76 [4]
4 years ago
12

i don't get it how the question says select all that apply and only give us one answer don't you think its common sense that the

re is more than one answer ???
Chemistry
1 answer:
Klio2033 [76]4 years ago
8 0
You should probably ask your teacher the next day, there may be a mistake.
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Curious Carl and his lab partner were handed a 2 liter sealed flask containing two gases, neon and argon. The partial of each ga
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Calculate the percent ionization of nitrous acid in a solution that is 0.222 M in nitrous acid (HNO3) and 0.278 M in potassium n
RideAnS [48]

Explanation:

Concentration of KNO_{2} is 0.278 M and it is completely ionized into K^{+} and NO^{-}_{2}.

This means that [KNO_{2}] = [NO_{2}] = 0.278 M

It is given that concentration of HNO_{2} is 0.222 M.

As HNO_{2} is a weak acid. Therefore, its dissociation will be as follows.

              HNO_{2}(aq) \rightarrow H^{+}(aq) + NO^{-}_{2}(aq)

Initially :    0.222 M            0        0.278 M

Change :    - x                    +x           +x

Equilibrium : 0.222 M - x    x         0.278 M + x

Therefore, dissociation constant for this reaction will be as follows.

        K_{a} = \frac{[H^{+}][NO^{-}_{2}]}{[HNO_{2}]}

Hence, putting the given values into the above formula as follows.

        K_{a} = \frac{[H^{+}][NO^{-}_{2}]}{[HNO_{2}]}

      4.50 \times 10^{-4} = \frac{x \times (0.278 + x)}{(0.222 - x)}

                          x = 0.00036

As, [H^{+}] is 0.00036 M. Therefore, percentage ionization of HNO_{2} will be calculated as follows.

              % ionization of HNO_{2} = \frac{[H^{+}]}{[HNO_{2}]} \times 100

                               = \frac{0.00036 M}{0.222 M} \times 100

                               = 0.16 %

Thus, we can conclude that % ionization of HNO_{2} is 0.16%.

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4 years ago
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