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kakasveta [241]
3 years ago
11

Sodium chloride (NaCl) is the chemical name for table salt and potassium chloride (KCl) is a common salt substitute. Using the p

eriodic table, choose the correct statement about how these compounds form. A. Both sodium and potassium atoms gain electrons to form ions. B. Sodium and potassium lose electrons and chlorine gains them. C. Chlorine gains electrons from potassium and loses electrons to sodium. D. Chlorine gains more electrons from potassium than it does from sodium.
Chemistry
1 answer:
Lina20 [59]3 years ago
7 0

Answer:

Option B. Sodium and potassium lose electrons and chlorine gains them

Explanation:

Sodium and Potassium are both group 1 element. They both react with chlorine by losing 1 electron each and the chlorine atom accept the electron.

Sodium chloride(NaCl) can be formed by the equation below:

Na —> Na+ + e-

Cl + e- —> Cl-

Combining both equation together we obtained the following:

Na + Cl + e- —> Na+ + Cl- + e-

Eliminating the e-, we have:

Na + Cl —> Na+Cl-

From the above illustration, we can see that sodium loses electron and chlorine accepts/gain the electron

This is also applicable to potassium chloride as shown below:

K —> K+ + e-

Cl + e- —> Cl-

Combining both equation together we obtained the following:

K + Cl + e- —> K+ + Cl- + e-

Eliminating the e-, we have:

K + Cl —> K+Cl-

we can see also that potassium loses electron and chlorine accepts/gain the electron

Therefore, the correct answer to the question is option B

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The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

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HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

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Equilibrium constant, Ka = 5.71×10¯⁷

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pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

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