Question

Calculate the concentration (M) of sodium ions in a solution made by diluting 25.0 mL of a 0.765 M solution of sodium sulfide to a total volume of 225 mL.

Answer 1

The concentration of sodium ions in the diluted ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}$ solution is $\boxed{{\text{0}}{\text{.17 M}}}$.

Further Explanation:

The concentration is the proportion of substance in the mixture. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The molarity equation for the dilution of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}$ solution is as follows:

${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}={{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}$                                      ......(1)

Here,

${{\text{M}}_{\text{1}}}$ is the molarity of the initial   solution.

${{\text{V}}_{\text{1}}}$ is the volume of the initial  solution.

${{\text{M}}_{\text{2}}}$ is the molarity of the diluted   solution.

${{\text{V}}_{_{\text{2}}}}$ is the volume of the diluted   solution.

Rearrange equation (1) to calculate ${{\text{M}}_{\text{2}}}$.

${{\text{M}}_{\text{2}}}=\dfrac{{{{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{V}}_{\text{2}}}}}$                          ......(2)

The value of ${{\text{M}}_{\text{1}}}$ is 0.765 M.

The value of ${{\text{V}}_{\text{1}}}$ is 25 mL.

The value of ${{\text{V}}_{_{\text{2}}}}$ is 225 mL.

Substitute these values in equation (2).

$\begin{aligned}{{\text{M}}_{{\text{HBr}}}}&=\frac{{\left( {{\text{0}}{\text{.765 M}}}\right)\times\left( {{\text{25 mL}}}\right)}}{{\left( {{\text{225 mL}}} \right)}}\\&= 0.085{\text{ M}}\\\end{aligned}$

Dilution is the conversion of a concentrated solution into a dilute solution with the addition of extra solvent but the amount of solute is unaltered. The change that arises is an increase in the volume of the solution.

The dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}$ occurs as follows:

${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}\rightleftharpoons {\text{2N}}{{\text{a}}^ + } + {{\text{S}}^{2 - }}$

According to the balanced chemical equation, one mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}$ dissociates to form two moles of ${\text{N}}{{\text{a}}^ + }$ . So the concentration of ${\text{N}}{{\text{a}}^ + }$ can be calculated as follows:

$\begin{aligned}{\text{Concentration of N}}{{\text{a}}^ + }&= 2\left( {{\text{0}}{\text{.085 M}}} \right)\\&=0.1{\text{7 M}}\\\end{aligned}$

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, Na+, Na2S, M1, M2, V1, V2, 0.17 M, dilution, 25 mL, 225 mL, 0.765 M, 0.085 M, molarity equation.

Answer 2

The concentration of sodium ion, Na⁺ in the solution made by diluting 25 mL of a 0.765 M solution of sodium sulfide, Na₂S, to a total volume of 225 mL is 0.17 M

• We'll begin by calculating the Molarity of the diluted solution of Na₂S. This can be obtained as follow:

Volume of stock solution (V₁) = 25 mL

Molarity of stock solution (M₁) = 0.765 M

Volume of diluted solution (V₂) = 225 mL

Molarity of diluted solution (M₂) =?

M₁V₁ = M₂V₂

0.765 × 25 = M₂ × 225

19.125 = M₂ × 225

Divide both side by 225

M₂ = 19.125 / 225

M₂ = 0.085 M

• Finally, we shall determine the concentration of sodium ion, Na⁺ in the diluted solution. This can be obtained as follow:

Na₂S(aq) —> 2Na⁺(aq) + S²¯(aq)

From the balanced equation above,

1 mole of Na₂S contains 2 moles of Na⁺

Therefore,

0.085 M Na₂S will contain = 2 × 0.085 = 0.17 M Na⁺

Thus, the concentration of sodium ion, Na⁺ in the diluted solution is 0.17 M

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