Chemistry, 29.02.2020 01:36 kcombest7219

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

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Answer from: kaylorb8284

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

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

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

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

Explanation:

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:

For a:

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.

For b:

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.

For c:

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.

For d:

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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