Can you please tell me if these are right?
Determine the pH for the following solutions
A [OH-] = 1.0 X 10^-7 M
I got 7
B. [H3O+ = 4.2 X 10 ^-3M
I got 2.4
C. [H3O+]= 0.0001M
I got 4
D. [OH-] = 8.5 X10^-9M
I got 5.9
What are the [H3O+] and [OH-] for a solution with the following pH values?
a. 10.0
For [OH-] I got 1.0*10^-4. and [H3O+] a. 10.0 I get 1.0*10^-10
Thank you!
And could you please tell me how to figure this question out?
Solution X has a pH of 9.5 and solution Y has a pH of 7.5.
a. what solution is more acidic?
b. what is the [H3O+] in each?
c. what is the [OH-] in each?
The first four are correct.
lower pH is more acid.
b. [H3O+]=log^-1 (9.5)=3.16E9
the antilog or log^-1 key might be the 10x key on your calculator.
pOH=14-pH=14-9.5=4.5
[OH-]=antilog (4.5)=3.2E4
Notice [OH][H}= 1E14, which offers another way to solve it.
To determine the pH for the given solutions, you need to use the formula:
pH = -log[H3O+]
A. For [OH-] = 1.0 X 10^-7 M, you can use the fact that [H3O+] * [OH-] = 1.0 X 10^-14 (water's auto-ionization constant at 25 degrees Celsius) to find the [H3O+] concentration.
[H3O+] = 1.0 X 10^-14 / [OH-]
[H3O+] = 1.0 X 10^-14 / (1.0 X 10^-7)
[H3O+] = 1.0 X 10^-7
Now you can calculate the pH using the formula:
pH = -log(1.0 X 10^-7)
pH = 7
Therefore, your answer is correct.
B. For [H3O+] = 4.2 X 10^-3 M, you already have the concentration of [H3O+]. You can directly calculate the pH using the pH formula:
pH = -log(4.2 X 10^-3)
pH ≈ 2.38
Your answer seems to have rounded to 2 decimal places, so it is approximately correct.
C. For [H3O+] = 0.0001 M, you can use the pH formula:
pH = -log(0.0001)
pH = 4
So, your answer is correct.
D. For [OH-] = 8.5 X 10^-9 M, you can use the fact that [H3O+] * [OH-] = 1.0 X 10^-14 to find [H3O+].
[H3O+] = 1.0 X 10^-14 / [OH-]
[H3O+] = 1.0 X 10^-14 / (8.5 X 10^-9)
[H3O+] ≈ 1.18 X 10^-6
Now you can calculate the pH:
pH = -log(1.18 X 10^-6)
pH ≈ 5.93
Your answer seems to have rounded to 2 decimal places, so it is approximately correct.
For the next question, to find [H3O+] and [OH-] for a given pH value, you can use the formula:
[H3O+] = 10^(-pH)
[OH-] = 1.0 X 10^(-14) / [H3O+]
a. For pH = 10.0:
[H3O+] = 10^(-10.0) = 1.0 X 10^(-10)
[OH-] = 1.0 X 10^(-14) / (1.0 X 10^(-10)) = 1.0 X 10^(-4)
Your answer for [OH-] is correct.
b. For pH = 10.0:
[H3O+] = 10^(-10.0) = 1.0 X 10^(-10)
c. For pH = 10.0, you've already calculated [OH-] = 1.0 X 10^(-4).
Now, let's solve the last question:
Solution X has a pH of 9.5, and solution Y has a pH of 7.5.
a. To determine which solution is more acidic, you can compare their pH values. The lower the pH, the more acidic the solution. In this case, Solution Y with a pH of 7.5 is more acidic than Solution X with a pH of 9.5.
b. To find [H3O+] in each solution:
[H3O+] for Solution X = 10^(-pH) = 10^(-9.5)
[H3O+] for Solution Y = 10^(-pH) = 10^(-7.5)
c. To find [OH-] in each solution, you can use the fact that [H3O+] * [OH-] = 1.0 X 10^(-14) and solve for [OH-]:
[OH-] for Solution X = 1.0 X 10^(-14) / [H3O+ for Solution X]
[OH-] for Solution Y = 1.0 X 10^(-14) / [H3O+ for Solution Y]
By calculating [H3O+] and [OH-], you can determine their values for each solution.
I hope this explanation helps you understand how to find pH, [H3O+], and [OH-] values. If you have any further questions, feel free to ask.