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what is the [H3O^+] in a solution in which the pOH=4.50?

**Answers and Views:**

*Answer by hotttiegrl9*

pOH = -log[OH-]

4.50 = -log[OH-]

so you do the anti log of -4.50

would be 10^-4.50

then that answer is [OH-]

1.0e-14 = [H3O+][OH-]

1.0e-14/[OH-] = [H3O+]

you just have to do the calculation to find [OH-] on the calculator and plug it in to the next equation.

hope this helps!

*Answer by Miroku*

Answer: 10^-9.50

Remember, H3O^+ is another way of writing H+.

Now remember that one formula in your book:

[H+][OH-] = [H2O] = 10^-14 (VERY important, remember this always)

[H+] = concentration (moles/liters) of H+, like wise with [OH-] and [H2O]

Here are the steps: Find pH, then convert pH into [H+]

To find pH, we have to know yet, another formula:

pH + pOH = 14 (also REMEMBER this, very important)

We have pOH, we can plug it in:

pH + 4.50 = 14, now solve for pH

14 – 4.50 = pH

pH = 9.5

Now, we have the pH, and want to find the [H+] which is the CONCENTRATION (note that pH is a measurement of how easily an H+ is able to transfer from an acid to a base).

Now, the last step, we have to yet know another formula:

pH = -log[H+] and pOH = -log[OH-] (REMEMBER)

To find [H+] while knowing pH = 9.50, we plug and solve this equation:

pH = -log[H+], plug in 9.50 for pH, we get:

9.50 = -log[H+], move the negative to the other side:

-9.50 = log[H+]

Take the anti-log (inverse log) of both sides (raise each side to the power of 10, know that 10 raise to a log cancels out the each other):

10^-9.50 = 10^log[H+]

10^-9.50 = [H+]

[H+] = [H30+] = 10^-9.50

Does that help? Good luck.

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