predict if precipitation will occur when 36ml of 0.0039M Na2CrO4 is mixed with 18ml of 0.00033M AgNO3.
M.W of Ag2CrO4 :331.73g/mol
Kspof ag2cro4 1.1x10^-12
To predict if precipitation will occur when two solutions are mixed, we need to compare the ion-product constant (Ksp) for the precipitate with the actual ionic product.
In this case, when Na2CrO4 and AgNO3 are mixed, the following reaction takes place:
2 AgNO3 + Na2CrO4 → Ag2CrO4(s) + 2 NaNO3
From the balanced equation, we can determine that 2 moles of silver nitrate (AgNO3) reacts with 1 mole of sodium chromate (Na2CrO4) to produce 1 mole of silver chromate (Ag2CrO4) precipitate.
First, we need to calculate the number of moles of silver chromate that can form. We'll start by calculating the number of moles of Na2CrO4 and AgNO3:
Moles of Na2CrO4 = Volume (in liters) x Molarity
= 36 ml / 1000 ml/L x 0.0039 moles/L
= 0.0001404 moles
Moles of AgNO3 = Volume (in liters) x Molarity
= 18 ml / 1000 ml/L x 0.00033 moles/L
= 0.00000594 moles
Since the Ag2CrO4 precipitate forms in a 1:1 ratio with AgNO3, we can conclude that no excess AgNO3 will be left after the reaction. Therefore, the number of moles of Ag2CrO4 that can potentially form is also 0.00000594 moles.
Next, we calculate the ionic product (IP) of Ag2CrO4 using the given Ksp value:
IP = [Ag+]^2 [CrO4^2-]
= (0.00000594 moles/L)^2
To determine if precipitation will occur, we need to compare the IP with the Ksp value:
If IP > Ksp, precipitation will occur.
If IP < Ksp, no precipitation will occur.
Plugging in the values:
(0.00000594 moles/L)^2 > 1.1 x 10^-12
Simplifying:
0.0000000353136 > 1.1 x 10^-12
Since 0.0000000353136 is greater than 1.1 x 10^-12, we can conclude that the ionic product exceeds the Ksp value. Therefore, precipitation of Ag2CrO4 will occur when 36 ml of 0.0039 M Na2CrO4 is mixed with 18 ml of 0.00033 M AgNO3.