Question

0.500 l of solution a, a metal nitrate solution that is 14.94% nitrogen by mass has same mass of solutes as 0.500 L of solution b, which contains an ionic compound consisting of potassium, hydrogen, and oxygen. When solution a

and b mixed and react completely 135.2 g of blue precipitate solution form. Starting solutions a and b identify all compounds and their concentrations

Answer 1

Hi there, blue precipitate can give by copper hydroxide as other compound consisting of potassium, hydrogen, and oxygen (KOH)

a metal nitrate solution that is 14.94% nitrogen by mass

Balanced equation: Cu(NO₃)₂ + 2 KOH => Cu(OH)₂ + 2 KNO₃

moles present in blue precipitate = 135.2 g / 97.56068 g/mol = 1.38580419899 mol

1 mol Cu(NO₃)₂ reacts with 2 mol KOH and produces 1 mol Cu(OH)₂

moles Cu(NO₃)₂ reacted = 1.38580419899 mol

mass of Cu(NO₃)₂ = 1.38580419899 mol x 187.5558 g/mol = 259.9 g

moles KOH reacted = 2 x 1.38580419899 mol = 2.77160839797 mol

mass of KOH = 2.77160839797 mol x 56.10564 g/mol = 155.5 g

concentration = moles / liter

[Cu(NO₃)₂] = 1.38580419899 mol / 0.5 L = 2.77160839797 M ~ 2.77 M

[KOH] = 2.77160839797 mol / 0.5 L = 5.54321679595 M ~ 5.54 M

Answers:

solution a, a metal nitrate = Cu(NO₃)₂ and its concentration = 2.77 M

solution b, ionic compound = KOH and its concentration = 5.54 M

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