Question

A 0.100 l solution of 0.300 m agno3 is combined with a 0.100 l solution of 1.00 m na3po4. calculate the concentration of ag and po43– at equilibrium after the precipitation of ag3po4 (ksp = 8.89 × 10–17).

Answer 1

moles of Ag+ = 0.1 L * 0.3 m = 0.03 Moles

moles of PO4-3 = 0.1 L * 1 = 0.1 moles


According to the reaction equation which can be written as:

                 3Ag +    +   PO4 3-   ↔   Ag3PO4(s)
initial        0.03                0.1
change    -3X                    -X                        -

Equ         (0.03-3X)      (0.1-X)

when Ksp = [Ag]^3 [PO43-]
               
                 = (0.03-3X)^3 ( 0.1-X)

∴8.89 x 10^-17 = (0.03 - 3X)^3 (0.1-X)

8.89 x 10^-17 = 27 * (X-0.1)^4

∴ X = 2.9 x 10^-6

∴[Ag+] = 0.03 - (3 * (2.9 X 10^-6))
  
             = 0.02998 M

[PO43-] = 0.1 - X = 0.1 - (4.3 x 10^-5)
                         
                             = 0.0999 M

Answer 2

The concentration of $\bold { Ag^+}$ is 0.02998 M while  the concentration of $\bold { PO_4^3^-}$  is 0.0999 M.

The reaction,

$\rm \bold{3Ag ^+ + PO_4^3^- \leftrightarrow Ag_3PO_4(s)}$

At equilibrium,

Concentration of $\bold { Ag^+}$   will be  (0.03-3X)

Concentration of $\bold { PO_4^3^-}$ will be  ( 0.1-X)

The moles of $\bold { Ag^+}$ is 0.03 Moles

Moles of $\bold { PO_4^3^-}$ is 0.1 moles

Solubility product constant formula,

$\rm \bold{ Ksp = [ Ag^+]^3[ PO_4^3^- ]}$

Put the values in the formula,Solve it for X

$\bold{8.89 x 10^-^1^7 = (0.03 - 3X)^3 (0.1-X)}$

$\bold {X = 2.9\times 10^-^6}$

Put the value of the X, in Ksp formula, we  get

$\rm \bold{ [ Ag^+] = 0.02998 M}\\\\\\rm \bold{ [PO_4^3^-] = 0.0999 M }$

Hence we can conclude that the concentration of $\bold { Ag^+}$ is 0.02998 M while  the concentration of $\bold { PO_4^3^-}$  is 0.0999 M.

To know more about Ksp, refer to the link:

https://brainly.com/question/17117417?referrer=searchResults