At 40 degrees Celsius, the value of Kw is 2.92 X 10^-14
a.) calculate the [H+] and [OH-] in pure water at 40 degrees celsius
b.)what is the pH in pure water at 40 degrees celsius
c.)if [OH-] is .18M , what is the pH
H2O ==> H^+ + OH^-
You know Kw = (H^+)(OH^-) = 2.92 x 10^-14.
AND you know (H^+) = (OH^-)in pure water.
Then pH = - log (H^+)
Post your work if you get stuck.
a. 1.71 x 10^-7
b.6.76
c. 12.54
(a)
H2O ==> H^+ + OH^-
for every x mols H2O that dissociate, there are x mols H^+ and x mols OH^- so
since Kw = (H^+)(OH^-) = 2.92 x 10^-14
then x*x= 2.92 x 10^-14
so x = sqrt 2.92 x 10^-14
(b)I assume you can do pH = -log(H^+) now that you know (H^+).
The answer you should obtain is 6.76731 which rounds to 6.77
Here is how you find pH from (H^+).
Suppose pH = 5.32
pH = - log(H^+)
5.32 = -log(H^+)
-5.32 = log(H^+)
So you take the antilog of -5.32. To do that, enter 5.32 on your calculator, change the sign to - (or enter -5.32 at the beginning), then punch the 10x key on your calculator. If you have done it right, you should get 4.7863 x 10^-6. Of course that's too many significant figures; however, I copied ALL of the digits so you can check your calculator ability. Now just to make sure you get it, use pH = 6.76731 and see if you get the same answer as you have for (a).
Let me know if this isn't clear.
I already worked out (c), but the rest is still just mush. I'm sorry for my stupidity, but this just doesn't make sense. I can find [H+] or [OH-] as long as I have the concentration of 1 of them, so, ???
By the way, how do you find H+ and OH- concentrations from the pH? I have scoured my text book, but found nothing.
a.) Well, if the value of Kw at 40 degrees Celsius is 2.92 x 10^-14, we can use that to find the [H+] and [OH-] in pure water. Since pure water is neutral, the concentration of [H+] and [OH-] will be the same.
So, let's call the concentration of both [H+] and [OH-] as 'x'.
Using the equation for Kw: Kw = [H+][OH-]
We can plug in the value of Kw and set it equal to 'x' squared: 2.92 x 10^-14 = x^2
To solve for 'x', we take the square root of both sides: x = sqrt(2.92 x 10^-14)
Calculating that results in a value for 'x' equal to 1.71 x 10^-7.
So, the concentration of both [H+] and [OH-] in pure water at 40 degrees Celsius is 1.71 x 10^-7 M.
b.) To find the pH in pure water, we can use the equation pH = -log[H+].
Since [H+] is equal to 1.71 x 10^-7 M, we plug that value into the equation: pH = -log(1.71 x 10^-7)
Calculating that results in a pH of approximately 6.767.
c.) Now, if [OH-] is 0.18 M, we can find the [H+] in the same way as before.
Using the equation for Kw: Kw = [H+][OH-]
We already know [OH-] is 0.18 M, so let's call the concentration of [H+] as 'y'.
Plugging in the values and solving for 'y': 2.92 x 10^-14 = (y)(0.18)
Solving for 'y' results in a value of 1.62 x 10^-13 M for [H+].
To find the pH, we use the equation: pH = -log[1.62 x 10^-13]
Calculating that results in a pH of approximately 12.79.
Hope that helps!
To solve these questions, we need to use the concept of the ion product of water, also known as Kw. The value of Kw changes with temperature. At 25 degrees Celsius, the value of Kw is 1.0 x 10^-14.
However, in this case, we are given that the value of Kw at 40 degrees Celsius is 2.92 x 10^-14. Let's use this information to answer the questions.
a.) To calculate the [H+] and [OH-] in pure water at 40 degrees Celsius, we can use the equation Kw = [H+][OH-]. Since pure water is neutral, [H+] and [OH-] have the same value.
Let's assume [H+] = [OH-] = x (Molarity).
Since Kw = 2.92 x 10^-14, we can write the equation as:
2.92 x 10^-14 = x * x
2.92 x 10^-14 = x^2
x = √(2.92 x 10^-14) ≈ 1.71 x 10^-7
So, at 40 degrees Celsius, the [H+] and [OH-] in pure water are approximately 1.71 x 10^-7 M.
b.) The pH is a measure of the hydrogen ion concentration. It is defined as the negative logarithm of [H+]. Mathematically, pH = -log[H+].
Using the value of [H+] we calculated in part a, the pH can be calculated as:
pH = -log(1.71 x 10^-7)
pH ≈ 6.77
Therefore, the pH of pure water at 40 degrees Celsius is approximately 6.77.
c.) If [OH-] is 0.18 M, we can use the equation Kw = [H+][OH-] to find the [H+] and then calculate the pH.
Let's assume [H+] = x (Molarity).
Kw = [H+][OH-]
2.92 x 10^-14 = x * 0.18
x = (2.92 x 10^-14) / 0.18 ≈ 1.62 x 10^-13
Now, using the value of [H+], we can calculate the pH:
pH = -log(1.62 x 10^-13)
pH ≈ 12.79
Therefore, if [OH-] is 0.18 M, the pH is approximately 12.79.