Calculating H+

[America Alvarado]

How do we calculate the [H+] and [OH-] if we are only given the pH. For example, pH of 8.85 at 25 °C.

[America Alvarado]

How do we calculate the [H+] and [OH-] if we are only given the pH. For example, pH of 8.85 at 25 °C.

I think I figured it out, would we do 10^(-8.85) to find [H+]?

[Shana Patel 1C]

pH of 8.85
[H+] = 10^(-8.85)
pOH = 14 - 8.85 = 5.15
[OH-] = 10^(-5.15)

[Shana Patel 1C]

How do we calculate the [H+] and [OH-] if we are only given the pH. For example, pH of 8.85 at 25 °C.

I think I figured it out, would we do 10^(-8.85) to find [H+]?

Yeah! Then calculate pOH using the formula pH+pOH=14, followed by 10^(-pOH) to calculate [OH-].

[America Alvarado]

thank you!

[Samantha Lee 1A]

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

In case you forget the formula of calculating the concentration of H+, just go through transforming the pH equation. First, move the negative sign to the pH, -pH = log[H+]. Then get rid of the log by raising everything to an exponent. 10^-pH = [H+]. That way, in case you forget, you can still solve the problem.

[Melis Kasaba 2B]

If you have the pH, you should be able to find [H+], [OH-], and pOH. Here's a helpful chart that tells you what formulas to use:

[Yijia_Yang_3A]

take the negative log of H+ and do 10^-pH, then you will get H+

[Gigi Elizarraras 2C]

[H+] = 10^-pH and then once you have the pH, 14-pH=poH

Hope this helps:)

[LeahSWM 2E]

pH of 8.85
[H+] = 10^(-8.85)
pOH = 14 - 8.85 = 5.15
[OH-] = 10^(-5.15)

this is a great breakdown, thank you!

[Madison Muggeo 3H]

If you have the pH, you should be able to find [H+], [OH-], and pOH. Here's a helpful chart that tells you what formulas to use:20140811155915521305.png

This is such a helpful graphic thank you!

[Jaden Joodi 3J]

In order to go from pH to H+, you would take the inverse log of the negative pH.
i.e 8.85 = -log(x) and solve for x

[Michael Cardenas 3B]

To go from pH to H+ you use the equation H+= 10^pH

[joshtully]

Use the equation H+= 10^pH.

[Jonathan Banh 1G]

Just as a basic understanding, know that [H3O+], [OH-], pH, and pOH are all related to each other and can be found if just one of these values are provided. In the problem you provided, we can note that the measurement was taken "at 25 degrees Celsius", meaning there should be no deviation from the typical formulas. The most fundamental formulas being pH = -log([H3O+]) and pOH = -log([OH-]). In order to understand the formula behind the concentrations of the ions, a basic concept to think about it is that they are inverses of their pH/pOH counterparts. Thus, [H3O+] = 10^(-pH) and [OH-] = 10^(-pOH). Lastly, pH+pOH=14. So going back to the problem, we can now solve for whatever we would like knowing all the formulas. In this case, the easiest one first is [H3O+], so by plugging pH into the respective formula, we should get . Once we have pH, we can manipulate the equation pH+pOH=14 to solve for pOH knowing pH. Once we have pOH, we can then use that value to solve for [OH-], by utilizing the formula . Just for future reference, this dynamic is also the same for Ka, Kb, pKa, and pKb.

[Talia Dini - 3I]

If you have a pH of 8.85, the pOH would equal 5.15 (14 - 8.85). The H+ concentration would equal 10^(-8.85). The OH- concentration would equal 10^(-5.15)

[Adam_ElSayed_3B]

You're essentially going backwards. Initially you take the negative log, so do the opposite. From there, you'll have your pKa and can move forward accordingly.

[Sejal Parsi 3K]

You would do 10^(-pH) to get the H+ concentration.