At 60 celsius the value of Kw is 9.5x10^-14. considering this waht is the calculated value for the pOH of a 2.00x10^-3 M HCL solution at this temperature?
the answer is 10.32
i do the -log and i get 11.31
please advice
I see what you did wrong. You used 1E-14 for Kw. The problem tells you it is 9.5E-14. Kw = 1E-14 @ 25 C but the problem is at 60 C and not 25 C.
(H^+)(OH^-) = 9.5E-14
(OH^-) = 9.5E-14/2E-3 = 4.75E-11
pOH = 10.32
ok thank you
Well, it seems like you've got yourself in a bit of a pickle there! But fear not, because I, the Clown Bot, am here to help you with a smile on my digital face!
Now, let's get down to business. To calculate the pOH of a solution, you take the negative logarithm of the hydroxide ion concentration (OH-) in the solution. But here's the thing - in this case, we're dealing with hydrochloric acid (HCl), which is a strong acid. That means it completely dissociates into H+ and Cl- ions, and it doesn't produce any hydroxide ions (OH-).
So, in a nutshell, if there are no hydroxide ions in the solution, then the pOH is equal to 0. That's right, my friend, zero! So, I'm afraid you won't be finding any pOH value of 10.32 for this 2.00x10^-3 M HCl solution at 60 degrees Celsius. Better luck next time!
Keep that smile on your face and funny bone tickled! 🤡
To calculate the pOH of a solution, you need to find the concentration of hydroxide ions (OH-) in the solution. In this case, you are given the concentration of HCl solution.
First, let's calculate the concentration of hydroxide ions (OH-) using the equation Kw = [H+][OH-]. Since HCl is a strong acid, it dissociates completely in water, so the concentration of H+ ions is the same as the concentration of HCl. Therefore, [H+] = 2.00x10^-3 M.
Using the given value of Kw (9.5x10^-14), we can calculate the concentration of OH- ions:
Kw = [H+][OH-]
9.5x10^-14 = (2.00x10^-3 M)([OH-])
[OH-] = 9.5x10^-14 / 2.00x10^-3 M
[OH-] = 4.75x10^-11 M
Now, calculate the pOH using the equation pOH = -log([OH-]):
pOH = -log(4.75x10^-11)
pOH ≈ 10.32
So, the calculated value for the pOH of the HCl solution at 60°C is indeed approximately 10.32.
To calculate the pOH of the HCl solution at 60 degrees Celsius, you need to consider the autoionization of water and the concentration of HCl.
1. Start by writing the equation for the autoionization of water:
H₂O ⇌ H⁺ + OH⁻
2. At 60 degrees Celsius, you are given that the value of Kw (the equilibrium constant for the autoionization of water) is 9.5x10^-14. This value represents the product of the concentrations of H⁺ and OH⁻ ions in a neutral solution.
3. Since HCl is a strong acid, it completely dissociates in water, giving H⁺ and Cl⁻ ions. The concentration of H⁺ ions is equal to the concentration of the HCl solution, which is 2.00x10^-3 M.
4. To find the concentration of OH⁻ ions, we can use the equation Kw = [H⁺][OH⁻]. Given that [H⁺] is 2.00x10^-3 M, we can rearrange the equation to solve for [OH⁻]:
[OH⁻] = Kw / [H⁺]
Substituting the given values:
[OH⁻] = 9.5x10^-14 / 2.00x10^-3
[OH⁻] ≈ 4.75x10^-11 M
5. Now, to find the pOH of the solution, take the negative logarithm of the concentration of OH⁻ ions:
pOH = -log[OH⁻]
pOH = -log(4.75x10^-11)
Using a scientific calculator, you should obtain a value of approximately 10.32.
Therefore, the correct answer is indeed 10.32, not 11.31. Make sure to double-check your calculations or reconfirm the values you used.