Question

A 1.00L of Solution of 0.310M AgNO3 is combined witha 0.1L solution of 1.0M Na3PO43- Calculate the concentration of Ag+ and PO43- at equilibrium after precipitation of Ag3PO4. Ksp = 8.89x10^-17

Answer 1

The concentration of [Ag⁺] and [PO₄³⁻] is 0.4485M and 5.83 × 10⁻⁶M after precipitation

Step-by-step explanation:

The volume of the mixture formed by mixing 0.100L of AgNO₃ and 0.100L of Na₃PO₄

0.100L + 0.100L = 0.200L

The ionic equation of AgNO₃

`AgNO_3_((aq)) \to Ag^+__((aq))+NO^-_3_((aq))`

The number of moles of the Ag⁺ in the mixture is calculated by

Number of mole = Molarity × Volume

= 0.310 × 0.100

= 0.031mole

The ionic equation for the Na₃PO₄³⁻

`Na_3PO_4_((aq)) \to3Na^+_((aq))+PO_4^3^-_((aq))`

Number of mole = Molarity × Volume

= 1.0 × 0.100

= 0.100 mole

One mole of Ag⁺ react three mole of PO₄³⁻, the number of mole of PO₄³⁻ react with 0.031mole of Ag⁺ is

number of mole PO₄³⁻ =

`=(1)/(3) * 0.0310\\\\= 0.0103mole`

Total number of mole remaining is

0.100 mole - 0.0103 mole

= 0.0897 mole

The concentration of PO₄³⁻ is calculated as shown below

Molarity = number of mole / volume

Molarity = 0.0897 / 0.2

= 0.4485M

The concentration of Ag⁺ is calculated as shown below

`K_p = [Ag^+]^3[PO^3^-_4]`

`8.89 * 10^-^1^7= [Ag^+]^3(0.4485M)`

`[Ag^+]^3 = (8.89* 10^-^1^7)/(0.4485) \\`

`[Ag^+]^3 = 1.982 * 10^-^1^6`

`[Ag^+] = 5.83 * 10^-^6M`

The concentration of [Ag⁺] and [PO₄³⁻] is 0.4485M and 5.83 × 10⁻⁶M after precipitation

Answer 2

Po₄³⁻ = 0.45 moles

(Ag⁺) = 5.89 x 10⁻⁶ moles

Step-by-step explanation:

Given data,

ksp = 8.89 x 10⁻¹⁷

Molarity of AgNo₃ = 0.31M

Total number of moles of silver ions = Molarity x volume

= 0.310 x 0.100

= 0.031 moles

Total number of mole of Phosphate ions = 1M x 0.100L

= 0.100 moles

3.1 moles of silver ions react with 1 mole of phosphate ions

1 mole of silver ions will react with 1/3 moles of phosphate ions

0.031 moles of silver ions react = 1/3 x 0.031

= 0.01 moles

Total number of moles of phosphate ions remaining = 0.100 - 0.01

= 0.09 moles

Total volume = 0.100 + 0.100

= 0.2

Po₄³⁻ = Total number of moles / Total volume

=0.09 / 0.2

Po₄³⁻ = 0.45 moles

Ksp = (Ag⁺)³(Po₄³⁻)

8.89 x 10-17 = (Ag⁺)³ x 0.45

(Ag⁺)³ = 8.89 x 10⁻¹⁷/0.45

(Ag⁺)³ =19.75 x 10-17

(Ag⁺) = 5.89 x 10⁻⁶ moles