What is the Qsp when 59.0 mL of 2.00 multiplied by 10-4 M AgNO3 are mixed with 70.0 mL of 1.00multiplied by10-4 M Na2CrO4? Assume the volumes are additive. Ksp = 1.1 multiplied by 10-12.
To find the Qsp (the reaction quotient), you need to write out the balanced chemical equation for the reaction that occurs between AgNO3 and Na2CrO4. From the given substances, it appears to be:
AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3
Now, let's determine the concentrations of the ions involved in the reaction.
For AgNO3:
- Volume: 59.0 mL
- Concentration: 2.00 × 10^(-4) M
For Na2CrO4:
- Volume: 70.0 mL
- Concentration: 1.00 × 10^(-4) M
Since the volumes are additive, the total volume of the mixture is 59.0 mL + 70.0 mL = 129.0 mL. However, to use the concentrations in the reaction quotient, we need to convert the volumes to liters.
Convert the volumes to liters:
- Volume of AgNO3: 59.0 mL = 59.0 × 10^(-3) L
- Volume of Na2CrO4: 70.0 mL = 70.0 × 10^(-3) L
Now we can calculate the Qsp using the concentrations and the balanced chemical equation:
Qsp = [Ag2CrO4] × [NaNO3]^2
The concentration of Ag2CrO4 can be calculated by dividing the number of moles of Ag2CrO4 by the total volume in liters. Since the stoichiometry of Ag2CrO4 in the balanced equation is 1:1, the concentration of Ag2CrO4 is the same as the concentration of [Ag2CrO4].
[Ag2CrO4] = (moles of Ag2CrO4) / (total volume in liters)
Similarly, the concentration of NaNO3 is double the concentration of [NaNO3].
[NaNO3] = (2 × moles of NaNO3) / (total volume in liters)
Since Ksp is given as 1.1 × 10^(-12), we can compare the Qsp to Ksp to determine if a precipitate will form.
If Qsp < Ksp, no precipitate will form.
If Qsp > Ksp, a precipitate will form.
Now, you can plug in the given values and calculate the Qsp.