Question

What is the silver ion concentration in a solution prepared by mixing 415 mL of 0.391 M silver nitrate with 419 mL of 0.496 M sodium phosphate? The Ksp of silver phosphate is 2.8 × 10-18

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Answer 1

Given :

Volume of AgNO3 = 415 mL = 0.415 L

[AgNO3]= 0.391 M

Na3PO4 = 419 mL = 0.419 L , [Na3PO4] = 0.496 M

Ksp of Ag3PO4 = 2.8 E -18

Lets show reaction between sodium phosphate and silver nitrate

3AgNO3 (aq)+ Na3PO4 (aq) --- > Ag3PO4 (s) + 3 NaNO3 (aq)

Net ionic equation

3 Ag⁺ (aq) + PO4^3- (aq) --- > Ag3PO4(s)

Calculation of limiting reactant

To calculate limiting reactant we need to find moles of each reactants.

n AgNO3 = volume in L x molarity

= 0.415 M x 0.391 M

= 0.1622 mol

n Na3PO4 = 0.2078 mol

Number of moles of silver nitrate required to completely react with sodium phosphate

= 0.2078 mol sodium phosphate x 3 mol silver nitrate / 1 mol sodium phosphate

= 0.6335 mol sodium silver nitrate

Actually moles of silver nitrate present = 0.1622 mol so silver nitrate is limiting reactant and it will consumed completely.

Now we can find moles of silver phosphate formed.

n silver phosphate = n silver nitrate x 1 mol silver phosphate / 3 mol silver nitrate

= 0.1622 mol silver nitrate x 1 mol silver phosphate / 3 mol silver nitrate

= 0.054088 moles of silver phosphate

[silver phosphate ]= 0.054088 mol / volume of solution in L

= 0.054088 mol / (0.415 + 0.419 ) L

= 0.0645 M

We use ksp in order to get concentration of silver ions.

Ksp = [ Ag+]³ [PO4^3-]

2.8 E -18 = (3x)³ ( x )

x = 2.36 E-5

now concentration of Ag+ = 3 x = 3 * 2.36 E-5 = 7.0851 E – 5 M