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o-na [289]
3 years ago
13

What is the overall charge on the compound, sodium hydroxide? explain

Chemistry
1 answer:
Fiesta28 [93]3 years ago
7 0

Answer:

Overall, the electric charge of the compound {\rm NaOH} is 0.

Explanation:

Sodium hydroxide {\rm NaOH} is an ionic compound. This compound is made up of a large number of sodium ions {\rm Na^{+}} and hydroxide ions {\rm OH^{-}}.

As the superscripts suggest, each sodium ion {\rm Na^{+}} carries a charge of (+1) while each hydroxide ion {\rm OH^{-}} carries a charge of (-1).

The two types of ions are paired up at a one-to-one ratio within {\rm NaOH}. Charges on these ions would balance one another, such that the overall electric charge on the compound {\rm NaOH}\! would be 0.

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Explanation:

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What is the percent ionization of a monoprotic weak acid solution that is 0.188 M? The acid-dissociation (or ionization) constan
djyliett [7]

Answer: The percent ionization of a monoprotic weak acid solution that is 0.188 M is 3.59\times 10^{-4}\%

Explanation:

Dissociation of weak acid is represented as:

HA\rightleftharpoons H^+A^-

 cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.188 M and \alpha = ?

K_a=2.43\times 10^{-12}

Putting in the values we get:

2.43\times 10^{-12}=\frac{(0.188\times \alpha)^2}{(0.188-0.188\times \alpha)}

(\alpha)=3.59\times 10^{-6}=3.59\times 10^{-4}\%

Thus percent ionization of a monoprotic weak acid solution that is 0.188 M is 3.59\times 10^{-4}\%

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