Lead pipes were once used to transport drinking water and are still found in some older homes and city water systems. Suppose the lead concentration and pH of the water determined by lead (II) hydroxide (Ksp = 2.8x10^-16) dissolving to produce a saturated solution. What would the lead concentration (M) and pH of this water be?
I am confused. Does this question actually have anything to do with lead pipes or is it a simple solubility problem?
Pb(OH)2 <--> Pb2+ + 2 OH-
Ksp = [Pb2+]([OH-]^2) = (s)((2s)^2) = 4(s^3)
Solve for s, which is [Pb2+].
Calculate [OH-], which is 2s, and then solve for pOH and lastly pH.
Ah, lead pipes and solubility problems – two of my favorite topics to clown around with! Let's dive in and solve this puzzle!
First, we start with the solubility product expression for lead (II) hydroxide (Pb(OH)2):
Ksp = [Pb2+][OH-]^2
Since the problem states that the water is saturated with lead hydroxide, we can assume that the concentration of Pb2+ is equal to the solubility (s).
Now, we can plug in and simplify:
Ksp = (s)((2s)^2) = 4s^3
To find the lead concentration [Pb2+], we solve for s:
4s^3 = Ksp
s^3 = Ksp/4
s = (Ksp/4)^(1/3)
Okay, now that we've got the lead concentration, let's calculate the hydroxide concentration [OH-]:
[OH-] = 2s
And since we're all about pH fun, let's move on to the pOH calculation:
pOH = -log [OH-]
And finally, the pH is just a clown's leap away:
pH = 14 - pOH
So, my friend, with your values of Ksp being 2.8x10^-16, all you have to do is plug it into our juggling equations and watch the results bounce back!
Remember, I'm just here to tickle your funny bone while providing answers. Double-check the calculations to be sure!
This question is related to the solubility of lead (II) hydroxide, which can be used to determine the lead concentration and pH of water with lead pipes. Here's a step-by-step approach to solve the problem:
1. Write the balanced equation for the dissociation of lead (II) hydroxide:
Pb(OH)2 <--> Pb2+ + 2 OH-
2. Write the expression for the solubility product constant (Ksp):
Ksp = [Pb2+]([OH-]^2)
3. Substitute the concentrations of the lead ion and hydroxide ion with 's' in the expression for Ksp:
Ksp = (s)((2s)^2) = 4(s^3)
4. Solve for 's', which represents the concentration of Pb2+:
4(s^3) = Ksp
s^3 = Ksp / 4
s = (Ksp / 4)^(1/3)
5. Calculate [OH-], which is twice the concentration of Pb2+:
[OH-] = 2s
6. Determine pOH using the concentration of hydroxide:
pOH = -log[OH-]
7. Calculate pH using the pOH value:
pH = 14 - pOH
By following these steps, you will be able to determine the lead concentration (M) and pH of the water based on the solubility of lead (II) hydroxide.
This question involves calculating the lead concentration and pH of water based on the solubility of lead (II) hydroxide, Pb(OH)2. Although the question mentions lead pipes, it is primarily a solubility problem that does not specifically address lead pipes.
To start, the solubility product constant (Ksp) for Pb(OH)2 is given as 2.8x10^-16. Knowing this, we can set up the equilibrium equation for the dissociation of Pb(OH)2:
Pb(OH)2 ⇌ Pb2+ + 2OH-
The Ksp expression for this equilibrium is Ksp = [Pb2+][OH-]^2. Since Pb(OH)2 dissociates into one Pb2+ ion and two OH- ions, we can write the expression as Ksp = [Pb2+](2[OH-])^2 = [Pb2+](4[OH-]^2).
Next, we can substitute the concentration of Pb2+ as 's' and the concentration of OH- as '2s'. Now, the Ksp expression becomes:
Ksp = 4s^3
To solve for the lead concentration, [Pb2+], we can rearrange the equation and isolate 's':
s = (Ksp/4)^(1/3)
Now that we know the lead concentration, we can calculate the OH- concentration by substituting 's' into the equation for OH-: [OH-] = 2s.
To find the pOH, we can take the negative logarithm of the OH- concentration: pOH = -log[OH-]. Finally, the pH can be obtained by subtracting the pOH from 14 (pH + pOH = 14).
So, to summarize the steps:
1. Calculate the lead concentration ([Pb2+]): s = (Ksp/4)^(1/3)
2. Calculate the hydroxide concentration ([OH-]): [OH-] = 2s
3. Find the pOH: pOH = -log[OH-]
4. Calculate the pH: pH = 14 - pOH
Note that the exact values of [Pb2+], [OH-], pOH, and pH depend on the given Ksp value and the calculation of 's'.