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Chapter 4: Problem 34

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.0200 mole of sodium phosphate in 10.0 mL of solution b. 0.300 mole of barium nitrate in 600.0 mL of solution c. 1.00 g of potassium chloride in 0.500 L of solution d. 132 g of ammonium sulfate in 1.50 L of solution

Short Answer

Expert verified
a. Molarity of \(Na^+\) = 6.00 M, Molarity of \(PO_4^{3-}\) = 2.00 M b. Molarity of \(Ba^{2+}\) = 0.500 M, Molarity of \(NO_3^-\) = 1.00 M c. Molarity of \(K^+\) = 0.0268 M, Molarity of \(Cl^-\) = 0.0268 M d. Molarity of \(NH_4^+\) = 1.33 M, Molarity of \(SO_4^{2-}\) = 0.667 M

Step by step solution

01

a. 0.0200 mole of sodium phosphate in 10.0 mL of solution

Sodium phosphate dissociates as follows: \(Na_3PO_4 \rightarrow 3Na^+ + PO_4^{3-}\) First, convert mL to L: 10.0 mL = 0.010 L Now, calculate the molarity for each ion: Molarity of \(Na^+\) = (0.0200 moles of Na_3PO_4 × 3 moles of \(Na^+\) per mole of Na_3PO_4) / 0.010 L = 6.00 M Molarity of \(PO_4^{3-}\) = (0.0200 moles of Na_3PO_4 × 1 mole of \(PO_4^{3-}\) per mole of Na_3PO_4) / 0.010 L = 2.00 M
02

b. 0.300 mole of barium nitrate in 600.0 mL of solution

Barium nitrate dissociates as follows: \(Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-\) First, convert mL to L: 600.0 mL = 0.600 L Now, calculate the molarity for each ion: Molarity of \(Ba^{2+}\) = (0.300 moles of Ba(NO_3)_2 × 1 mole of \(Ba^{2+}\) per mole of Ba(NO_3)_2) / 0.600 L = 0.500 M Molarity of \(NO_3^-\) = (0.300 moles of Ba(NO_3)_2 × 2 moles of \(NO_3^-\) per mole of Ba(NO_3)_2) / 0.600 L = 1.00 M
03

c. 1.00 g of potassium chloride in 0.500 L of solution

Potassium chloride dissociates as follows: \(KCl \rightarrow K^+ + Cl^-\) First, convert 1.00 g of potassium chloride to moles (molecular weight of KCl = 39.1 + 35.5 = 74.6 g/mol): 1.00 g KCl × (1 mol KCl / 74.6 g KCl) = 0.0134 moles of KCl Now, calculate the molarity for each ion: Molarity of \(K^+\) = (0.0134 moles of KCl × 1 mole of \(K^+\) per mole of KCl) / 0.500 L = 0.0268 M Molarity of \(Cl^-\) = (0.0134 moles of KCl × 1 mole of \(Cl^-\) per mole of KCl) / 0.500 L = 0.0268 M
04

d. 132 g of ammonium sulfate in 1.50 L of solution

Ammonium sulfate dissociates as follows: \((NH_4)_2SO_4 \rightarrow 2NH_4^+ + SO_4^{2-}\) First, convert 132 g of ammonium sulfate to moles (molecular weight of (NH_4)_2SO_4 = (2 × 18.0) + 32.1 + (4 × 16.0) = 132.1 g/mol): 132 g (NH_4)_2SO_4 × (1 mol (NH_4)_2SO_4 / 132.1 g (NH_4)_2SO_4) = 1.00 moles of (NH_4)_2SO_4 Now, calculate the molarity for each ion: Molarity of \(NH_4^+\) = (1.00 moles of (NH_4)_2SO_4 × 2 moles of \(NH_4^+\) per mole of (NH_4)_2SO_4) / 1.50 L = 1.33 M Molarity of \(SO_4^{2-}\) = (1.00 moles of (NH_4)_2SO_4 × 1 mole of \(SO_4^{2-}\) per mole of (NH_4)_2SO_4) / 1.50 L = 0.667 M

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