For each solution, determine the p-values for each ion indicated. a solution that is 0.493 m in nacl and 0.314 m in nh 4 cl .

Complete Question:

Ions to calculate the p-values: Na⁺, Cl⁻, and NH₄⁺

pNa = 0.307

pCl = 0.093

pNH₄ = 0.503

Explanation:

The p-value is calculated by the antilog of the concentration of the substance of interest. For example, pH = -log[H⁺]. Thus, first, let's find the ions concentration.

Both substances are salts that solubilize completely, thus, by the solution reactions:

NaCl → Na⁺ + Cl⁻

NH₄Cl → NH₄⁺ + Cl⁻

So, for both reactions the stoichiometry is 1:1:1 and the concentration of the ions is equal to the concentration of the salts.

[Na⁺] = 0.493 M

[Cl⁻] = 0.493 + 0.314 = 0.807 M

[NH₄⁺] = 0.314 M

The p-values are:

pNa = -log[Na⁺] = -log(0.493) = 0.307

pCl = -log[Cl⁻] = -log(0.807) = 0.093

pNH₄ = -log[NH₄⁺] = -log(0.314) = 0.503

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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