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

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

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.