If a solution of 20mL of 0.050M K+ is added to 80mL of 0.50M ClO4- will a precipitate form and what is the value of Qsp? For KClO4, Ksp = 1.07 x 10-2. Thank you for any help! I don't know what I am doing in this problem!
I agree with Qsp but I disagree with the calculation to determine if a ppt will form or not. It isn't mols one wants; it is molarity.
M K^+ = 0.05 x (20/100) = ?
M ClO4^- = 0.5 x (80/100) = ?
Plug those concentrations into Qsp to determine if Qsp > Ksp.
Forgot a step : Solve for moles AND TAKE THE TOTAL VOLUME WHICH IS 0.1L AND CALCULATE MOLARITY. Caps is just to emphasize the part I forgot, not to yell.
you forgor 💀
KClO4 --------->. K^+ + ClO4^-
Qsp=[K^+][ClO4^-]
moles=Molarity*volume (L)
Solve for moles of each
If Qsp>Ksp, precipitate forms, if less, no precipitate forms.
To determine if a precipitate will form when the two solutions are mixed, we need to compare the value of Qsp (the ion product) with the Ksp (the solubility product constant).
To calculate Qsp, we first need to determine the initial number of moles of K+ and ClO4- ions present in each solution.
For the K+ solution:
moles of K+ = concentration (M) × volume (L)
= 0.050 M × 0.020 L
= 0.001 mol
For the ClO4- solution:
moles of ClO4- = concentration (M) × volume (L)
= 0.50 M × 0.080 L
= 0.040 mol
Now that we have determined the initial number of moles of each ion, we can write the balanced equation for the dissociation of KClO4 in water:
KClO4(s) ⇌ K+(aq) + ClO4-(aq)
Since 1 mole of KClO4 produces 1 mole of K+ and 1 mole of ClO4-, the number of moles will remain the same after dissociation.
Now, let's calculate the concentrations of K+ and ClO4- ions in the final solution:
For K+ ions:
final concentration (M) = moles of K+ / total volume (L)
= 0.001 mol / (0.020 L + 0.080 L)
= 0.01 M
For ClO4- ions:
final concentration (M) = moles of ClO4- / total volume (L)
= 0.040 mol / (0.020 L + 0.080 L)
= 0.40 M
Now we have the concentrations of K+ and ClO4- ions in the final solution. To determine if a precipitate will form, we compare the Qsp with the Ksp.
The Qsp is calculated by multiplying the ion concentrations:
Qsp = [K+][ClO4-]
= (0.01 M)(0.40 M)
= 0.004 M^2
The Ksp for KClO4 is given as 1.07 × 10^-2. Comparing Qsp with Ksp, we see that Qsp (0.004) is less than Ksp (1.07 × 10^-2).
Since Qsp < Ksp, no precipitate will form.
In summary, when 20 mL of 0.050 M K+ solution is added to 80 mL of 0.50 M ClO4- solution, a precipitate will not form and the value of Qsp is 0.004 M^2.