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3 years ago

Ionic equation for Sr3(PO4)2 and its solubility product (Ksp)?

Chemistry

1 answer:

Rus_ich [418]3 years ago

4 0

Answer:

$Sr_3(PO_4)_2(s)\rightleftharpoons 3Sr^{+2}(aq)+2(PO_4)^{-3}(aq)$

$Ksp=[Sr^{+2}]^3[(PO_4)^{-3}]^2$

Explanation:

Hello,

In this case, for strontium phosphate, we find an ionic equation for its dissociation as shown below:

$Sr_3(PO_4)_2(s)\rightleftharpoons 3Sr^{+2}(aq)+2(PO_4)^{-3}(aq)$

Next, the solubility product is found by applying the law of mass action, considering that the solid salt is not considered but just the aqueous species due to heterogeneous equilibrium:

$Ksp=[Sr^{+2}]^3[(PO_4)^{-3}]^2$

Best regards.

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Answer:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

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Note: The question is stated more clearly below:

Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.

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Number of moles in 100 g mass = % mass / molar mass

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1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

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Number of moles of each atom:

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Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

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Number of moles of each atom:

V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

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3 years ago

Ionic equation for Sr3(PO4)2 and its solubility product (Ksp)?​

Ionic equation for Sr3(PO4)2 and its solubility product (Ksp)?