Answer:
Po₄³⁻ = 0.45 moles
(Ag⁺) = 5.89 x 10⁻⁶ moles
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
Answer:
The concentration of [Ag⁺] and [PO₄³⁻] is 0.4485M and 5.83 × 10⁻⁶M after precipitation
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
[tex]AgNO_3_{(aq)} \to Ag^+__{(aq)}+NO^-_3_{(aq)}[/tex]
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⁻
[tex]Na_3PO_4_{(aq)} \to3Na^+_{(aq)}+PO_4^3^-_{(aq)}[/tex]
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⁻ =
[tex]=\frac{1}{3} \times 0.0310\\\\= 0.0103mole[/tex]
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
[tex]K_p = [Ag^+]^3[PO^3^-_4][/tex]
[tex]8.89 \times 10^-^1^7= [Ag^+]^3(0.4485M)[/tex]
[tex][Ag^+]^3 = \frac{8.89\times 10^-^1^7}{0.4485} \\[/tex]
[tex][Ag^+]^3 = 1.982 \times 10^-^1^6[/tex]
[tex][Ag^+] = 5.83 \times 10^-^6M[/tex]
The concentration of [Ag⁺] and [PO₄³⁻] is 0.4485M and 5.83 × 10⁻⁶M after precipitation
