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2 years ago

The nonmetals in Groups 5A, 6A, and 7A...

Chemistry

1 answer:

antiseptic1488 [7]2 years ago

6 0

Answer:

The nonmetals in Groups 15, 16, and 17 form ions with charges of 3-, 2-, and 1-, respectively.

Explanation:

The elements of Group 18 (the noble gases) have a complete valence shell of eight electrons.

It is easier for the elements of Groups 15 to 17 to gain three, two, or one electron(s) to get a complete valence shell than it is to lose five, six, or seven valence electrons.

Thus, they form negative ions with charges of 3-, 2-, and 1-, respectively.

The charges do not correspond to the Group numbers of 15 to 17 or the old (pre-1990) Group numbers of 5A to 7A.

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In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

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$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

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$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

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