What is the pH of a solution having a hydrogen ion concentration of 1x10^-5 M?
Also the same question, but Hydroxide instead of hydrogen.
thanks in advance!
Well, let me make you laugh before I give you the answer. Why don't chemists like bookstores? Because they have too many Alkynes of books! Now, getting down to business.
For the first question, the pH of a solution with a hydrogen ion concentration of 1x10^-5 M can be calculated using the formula: pH = -log[H+]. So, taking the negative logarithm of 1x10^-5, we get a pH of 5.
Now, let's move on to the second question. The concentration of hydroxide ions (OH-) can be used to calculate the pOH, and then we can determine the pH using pH + pOH = 14. However, since no hydroxide ion concentration was provided in the question, I'm afraid I can't calculate the pH for you. But hey, at least I made you smile, right?
To calculate the pH of a solution, you can use the formula: pH = -log[H+].
For the first question, where the hydrogen ion concentration is 1x10^-5 M, substitute the value in the formula:
pH = -log[1x10^-5]
pH = -log(0.00001)
Use a calculator to evaluate the logarithm:
pH ≈ -(-5)
pH ≈ 5
Therefore, the pH of the solution with a hydrogen ion concentration of 1x10^-5 M is approximately 5.
For the second question, if you have the hydroxide ion concentration and want to calculate the pOH, you can use the formula: pOH = -log[OH-].
However, to convert from pOH to pH, you can use the equation: pH + pOH = 14.
So for the hydroxide ion concentration, the pOH would be equal to the negative logarithm of the concentration.
If the hydroxide ion concentration is also 1x10^-5 M, the pOH can be calculated as:
pOH = -log[1x10^-5]
pOH = -log(0.00001)
Again, using a calculator to evaluate the logarithm:
pOH ≈ -(-5)
pOH ≈ 5
Since pH + pOH = 14, you can subtract the pOH from 14 to find the pH:
pH = 14 - pOH
pH = 14 - 5
pH = 9
Therefore, the pH of the solution with a hydroxide ion concentration of 1x10^-5 M is approximately 9.
To find the pH of a solution with a given hydrogen ion concentration, you can use the equation:
pH = -log[H+]
In this case, the hydrogen ion concentration is given as 1x10^-5 M. Plugging this value into the equation, you get:
pH = -log(1x10^-5)
To evaluate this expression, take the logarithm of the concentration and change the sign:
pH = -log(1x10^-5) = -(-5) = 5
So, the pH of the solution with a hydrogen ion concentration of 1x10^-5 M is 5.
Similarly, to find the pOH of a solution with a given hydroxide ion concentration, you can use the equation:
pOH = -log[OH-]
The hydroxide ion concentration can be related to the hydrogen ion concentration using the equation:
[OH-] x [H+] = 1x10^-14
In this case, you are given the hydrogen ion concentration as 1x10^-5 M. So, you can calculate the hydroxide ion concentration:
[OH-] = (1x10^-14) / [H+]
[OH-] = (1x10^-14) / (1x10^-5)
[OH-] = 1x10^-14 / 1x10^-5
[OH-] = 1x10^-14 x 1x10^5
[OH-] = 1x10^-9
Now, you can find the pOH using the formula:
pOH = -log(1x10^-9)
Evaluate this expression by taking the logarithm and changing the sign:
pOH = -log(1x10^-9) = -(-9) = 9
So, the pOH of the solution with a hydroxide ion concentration of 1x10^-9 M is 9.
You do these in your head.
pH = -log(H^+) = -log(1E-5) = 5
pOH then is 14-5 = 9
pH = -log(H^+)
pOH = -log(OH^-)
pH + pOH = pKw = 14
and (H^+)(OH^-) = Kw = 1E-14
These three will take care of any
pH/pOH/H^+/OH^- problems you will ever have.