Question

What is the concentration of each ion in a solution that is prepared by dissolving 5.00 g of ammonium chloride in enough water to make a 500.0 ml solution?

Answer 1

The concentration of each ion in the solution of ammonium chloride is:

• NH₄⁺ = 0.1886 M
• Cl⁻ = 0.1886 M

To solve this problem, the formulas and the procedures that we have to use are:

• M = n(solute)/v(solution) L
• n = m / MW
• MW= ∑ AWT

Where:

• M= molarity
• n = moles
• m = mass
• v = volume
• MW = molecular weight
• AWT = atomic weight

Information about the problem:

• m(NH₄Cl) = 5 g
• v(solution) = 500 ml
• AWT (N)= 14 g/mol
• AWT (H)= 1 g/mol
• AWT (Cl) = 35 g/mol

Converting the volume units from (ml) to (L) we have:

v(solution) = 500 ml * (1 L/1000 ml)

v(solution) = 0,50 L

We calculate the moles of the NH₄Cl from the MW:

MW = ∑ AWT

MW (NH₄Cl)= AWT (N) + AWT (H)*4 + AWT (Cl)

MW (NH₄Cl)= 14 g/mol +( 1 g/mol * 4) + 35 g/mol

MW (NH₄Cl)= 14 g/mol + 4 g/mol + 35 g/mol

MW (NH₄Cl)= 53 g/mol

Having the MW we calculate the moles of NH₄Cl:

n(NH₄Cl) = m(NH₄Cl) / MW(NH₄Cl)

n(NH₄Cl) = m(H2SO4) / MW (H2SO4)

n(NH₄Cl) =  5 g / 53 g/mol

n(NH₄Cl) = 0.0943 mol

Applying the molarity formula, we get:

M(NH₄Cl) = n(NH₄Cl)/v(solution) L

M(NH₄Cl) = 0.0943 mol / 0,50 L

M(NH₄Cl) = 0.1886 M

There are 0.1886 moles of NH₄Cl per liter of solution.

Let's recognize that 1 mol NH₄Cl contains:

• 1 mol NH₄⁺
• 1 mol Cl⁻

The concentration of each ion is thus:

(1)*(0.1886 M) = 0.1886 M NH₄⁺

(1)*(0.1886 M) = 0.1886 M Cl⁻

What is a solution?

In chemistry a solution is known as a homogeneous mixture of two or more components called:

• Solvent
• Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

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