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elixir [45]
3 years ago
5

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.a. 0.0200 mol sodium p

hosphate in 10.0 mL of solution. 0.300 mol of barium nitrate in 600.0 mL of solution. 1.00 g of potassium chloride in 0.500 L of solution. 132 g of ammonium sulfate in 1.50 L of solution
Chemistry
1 answer:
JulsSmile [24]3 years ago
4 0

<u>Answer:</u>

<u>For a:</u> The concentration of Na^+\text{ and }PO_4^{3-} ions in the solution are 6 M and 2 M respectively.

<u>For b:</u> The concentration of Ba^{2+}\text{ and }NO_3^{-} ions in the solution are 0.5 M and 1.0 M respectively.

<u>For c:</u> The concentration of K^{+}\text{ and }Cl^{-} ions in the solution are 0.051 M and 0.051 M respectively.

<u>For d:</u> The concentration of NH_4^{+}\text{ and }SO_4^{2-} ions in the solution are 1.34 M and 0.67 M respectively.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}       ......(1)

Or,

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}     ...(2)

For the given options:

  • <u>For a:</u>

The chemical formula of sodium phosphate is Na_3PO_4

Moles of sodium phosphate = 0.0200 moles

Volume of solution = 10.0 mL

Putting values in equation 1, we get:

\text{Molarity of the sodium phosphate}=\frac{0.0200\times 1000}{10.0}=2M

1 mole of sodium phosphate produces 3 moles of Na^+ ions and 1 mole of PO_4^{3-} ions

So, concentration of Na^+\text{ ions}=(3\times 2)=6M

Concentration of PO_4^{3-}\text{ ions}=(1\times 2)=2M

Hence, the concentration of Na^+\text{ and }PO_4^{3-} ions in the solution are 6 M and 2 M respectively.

  • <u>For b:</u>

The chemical formula of barium nitrate is Ba(NO_3)_2

Moles of barium nitrate = 0.300 moles

Volume of solution = 600.0 mL

Putting values in equation 1, we get:

\text{Molarity of the barium nitrate}=\frac{0.300\times 1000}{600.0}=0.5M

1 mole of barium nitrate produces 1 mole of Ba^{2+} ions and 2 mole of NO_3^{-} ions

So, concentration of Ba^{2+}\text{ ions}=(1\times 0.5)=0.5M

Concentration of NO_3^{-}\text{ ions}=(2\times 0.5)=1M

Hence, the concentration of Ba^{2+}\text{ and }NO_3^{-} ions in the solution are 0.5 M and 1.0 M respectively.

  • <u>For c:</u>

The chemical formula of potassium chloride is KCl

Given mass of potassium chloride = 1.00 g

Molar mass of potassium chloride = 39 g/mol

Volume of solution = 0.500 L

Putting values in equation 1, we get:

\text{Molarity of the potassium chloride}=\frac{1.00}{39\times 0.500}=0.051M

1 mole of potassium chloride produces 1 mole of K^{+} ions and 1 mole of Cl^{-} ions

So, concentration of K^{+}\text{ ions}=(1\times 0.051)=0.051M

Concentration of Cl^{-}\text{ ions}=(1\times 0.051)=0.051M

Hence, the concentration of K^{+}\text{ and }Cl^{-} ions in the solution are 0.051 M and 0.051 M respectively.

  • <u>For d:</u>

The chemical formula of ammonium sulfate is (NH_4)_2SO_4

Given mass of ammonium sulfate = 132 g

Molar mass of ammonium sulfate = 132 g/mol

Volume of solution = 1.50 L

Putting values in equation 1, we get:

\text{Molarity of the ammonium sulfate}=\frac{132}{132\times 1.50}=0.67M

1 mole of ammonium sulfate produces 2 moles of NH_4^{+} ions and 1 mole of SO_4^{2-} ions

So, concentration of NH_4^{+}\text{ ions}=(2\times 0.67)=1.34M

Concentration of SO_4^{2-}\text{ ions}=(1\times 0.67)=0.67M

Hence, the concentration of NH_4^{+}\text{ and }SO_4^{2-} ions in the solution are 1.34 M and 0.67 M respectively.

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