Q4.18P

Question

Question: How many moles and numbers of ions of each type are present in the following aqueous solutions?

(a) 130 mL of 0.45 M aluminum chloride

(b) 9.80 mL of a solution containing 2.59 g lithium sulfate/L

(c) 245 mL of a solution containing formula units of potassium bromide per liter.

Step-by-Step Solution

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Answer

Answer

The moles and numbers of ions of each type are present in the following aqueous solutions are:

(a) 0.059 mol AI³⁺ 0.176 mol CI^-

$355 \times 10^{22} A I^{3 +} i o n s, 1.06 \times 10^{23} C I^{-}^{}$

$(b) 4.62 \times 10^{- 4} m o l L i^{+}, 2.31 \times 10^{- 4} m o l s {o_{4}}^{2 -} 2.78 \times 10^{20} m o l L i^{+} i o n s, 1.39 \times 10^{20} m o l s {o_{4}}^{2 -} i o n s (c) 1.50 \times 10^{- 2} m o l K^{+}, 1.50 \times 10^{- 2} m o l {B_{r}}^{-}$

1Step 1: Calculate moles of ions released when dissolve in water

The dissociation reaction of aluminium chloride is:

$A I C I_{3} (s) \to A I^{3 +} (aq) + 3 C I^{-} (aq)$

The moles of $A I^{3 +}$ is calculated as:

$m o l e s o f A I^{3 +} = (130 mI) (\frac{1 L}{1000 ml}) (\frac{0.45 moI {AICI}_{3}}{Lit}) (\frac{1 {molH}^{+}}{1 mol {AICI}_{3}}) = 0.059 m o l A I^{3 +}$

The moles of $C I^{-}$ is calculated as:

$m o l e s o f A I^{3 +} = (130 mI) (\frac{1 L}{1000 ml}) (\frac{0.45 moI {AICI}_{3}}{Lit}) (\frac{3 m o l C I^{-}}{1 mol {AICI}_{3}}) = 0.1769 m o l C I^{-}$

2Step 2: Calculate number of ions of AI 3+ and CI -

The number of ions of AI³⁺ is:

$A I^{3 +} i o n s = (0.059 m o l A I^{3 +}) (\frac{6.022 \times 10^{23} A I^{3 +} i o n .}{1 m o l A I^{3 +}} = 3.55 \times 10^{22} m o l A I^{3 +} i o n s$

The number of ions of CI^-is:

$m o l e s o f C I^{-} = (0.1769 m o l C I^{-}) (\frac{6.022 \times 10^{23} A I^{3 +} i o n .}{1 m o l C I^{-}} = 1.06 \times 10^{23} m o l C I^{-} i o n s$

3Step 3: Calculate moles of ions released when dissolve in water

The dissociation reaction of lithium sulfate is:

$L i_{2} S O_{4} (s) \to 2 L I^{+} (aq) + S {O_{4}}^{2 -} (aq)$

The moles of $L i^{+}$ is calculated as:

$m o l e s o f L i^{+} = (9.80 mI) (\frac{1 L}{1000 ml}) (\frac{2.59 g {Li}^{2} S {O_{4}}^{}}{Lit}) (\frac{1 mole of {Li}^{+}}{109.95 g {Li}^{2} S {O_{4}}^{}}) = 4.62 \times 10^{- 4} L i^{+} i o n s$

The moles of $S {O_{4}}^{2 -}$ is calculated as:

$m o l e s o f S {O_{4}}^{2 -} i o n s = (9.80 m I) (\frac{1 L}{1000 m l}) (\frac{2.59 g {Li}^{2} S {O_{4}}^{}}{Lit}) (\frac{1 m o l e o f {Li}^{2} {SO}_{4}}{109.95 g {Li}^{2} S {O_{4}}^{}}) = 2.31 \times 10^{- 4} S {O_{4}}^{2 -} i o n s$

4Step 4: Calculate number of ions of S O 4 2 -   and Li +

The number of Li⁺ ions of is:

$L i^{+} i o n s = (4.62 \times 10^{- 4} m o l L i^{+}) (\frac{6022 \times 10^{23} C I^{-} i o n s .}{1 m o l C I^{-} i o n}) 2.78 \times 10^{20} L i^{+} i o n s$

The number of $S {O_{4}}^{2 -}$ ions of is:

$S {O_{4}}^{2 -} i o n s = (2.31 \times 10^{- 4} S {O_{4}}^{2 -} ions) (\frac{6022 \times 10^{23} S {O_{4}}^{2 -} i o n s .}{1 m o l S {O_{4}}^{2 -} i o n}) = 1.39 \times 10^{20} S {O_{4}}^{2 -} i o n s$

5Step 5: Calculate moles of ions released when dissolve in water

The dissociation reaction of potassium bromide is:

$K B r (s) \to K^{+} (aq) + B r^{-} (aq)$

The moles of K⁺is calculated as:

$k^{+} i o n s = (9.02 \times 10^{21} k^{+} ions) (\frac{1 m o l k^{+}}{6022 \times 10^{23} k^{+} i o n s .}) = 1.50 \times 10^{23} \times 10^{2} m o l k^{+}$

The moles of Br ^-is calculated as:

$B r i o n s = (9.02 \times 10^{21} {Br}^{-} ions) (\frac{1 m o l b r^{-}}{6022 \times 10^{23} B r i o n s .}) = 1.50 \times 10^{23} \times 10^{2} m o l B r^{-}$

6Step 6: Calculate number of ions of and

The number of K⁺ ions of is:

$K^{+} i o n s = (245 mI) (\frac{1 L}{1000 ml}) (\frac{3.68 \times 10^{22} F . U . KBr}{Lit}) (\frac{1 B r i o n s}{1 F . U . KBr}) = 9.02 \times 10^{21} K^{+} i o n s$

The number of B_r^-ions of is:

$B r i o n s = (245 mI) (\frac{1 L}{1000 ml}) (\frac{3.68 \times 10^{22} F . U . KBr}{Lit}) (\frac{1 B r i o n s}{1 F . U . KBr}) = 9.02 \times 10^{21} B r^{-} i o n s$

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