calculating pOH when given only [H+]
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calculating pOH when given only [H+]
How are we able to calculate the pOH when we're given only the concentration of H+ ions? Are we supposed to use the Kw of water or something like that?
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Re: calculating pOH when given only [H+]
Hello!
When given [H+], you can use the formula pH=-log[H+] to find pH. Once you find pH, you can subtract it by 14 (14-pH=pOH) to get pOH. Hope this helps!
When given [H+], you can use the formula pH=-log[H+] to find pH. Once you find pH, you can subtract it by 14 (14-pH=pOH) to get pOH. Hope this helps!
Re: calculating pOH when given only [H+]
pOH is derived from -log[OH-]
If you're given the [H+], you can find the [OH-] by diving 10^-14 by the [H+]. You can then take this calculated [OH-] and enter it into the above equation to calculate the pOH.
If you're given the [H+], you can find the [OH-] by diving 10^-14 by the [H+]. You can then take this calculated [OH-] and enter it into the above equation to calculate the pOH.
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Re: calculating pOH when given only [H+]
Hi, to calculate the pOH when only given the [H+] concentration, you first use that value to calculate the [OH-] value by dividing the [H+] value into 1.0 x 10^-14. From there, you take the quotient value of [OH-] and take the negative log of it to calculate the pOH. Hope this helps!
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Re: calculating pOH when given only [H+]
When you have [H+], you have to calculate either the pH or the [OH-] concentration before you can solve for pOH.
This is how you solve for pOH via the pH calculation.
From [H+], you can use pH = -log[H3O+] (or -log[H+], since these will be the same thing) to get the pH value.
Now, you can subtract the pH from 14 to get the pOH value since we know that pH + pOH = 14.
This is how you solve for pOH via the pH calculation.
From [H+], you can use pH = -log[H3O+] (or -log[H+], since these will be the same thing) to get the pH value.
Now, you can subtract the pH from 14 to get the pOH value since we know that pH + pOH = 14.
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Re: calculating pOH when given only [H+]
If you are given the concentration of H+, then you can calculate the pH using -log(H+) and from there, subtract this value from 14 to find the pOH. This is the case because pOH + pH = 14.
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Re: calculating pOH when given only [H+]
I had this same question during the achieve problems. I found that subtracting the -log(H+) aka the pH from 14 was the most efficient method.
Re: calculating pOH when given only [H+]
When calculating pOH when given the concentration of H+ ions, it is easier to find the pH by using the equation (pH=-log[H+]) and then subtracting the answer from 14, but you can also choose to go through the long process of dividing 1x10^-14 by [H+] and then taking the -log[OH-] (although I don't recommend using this method, it may help you better understand the concept of pH and pOH).
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Re: calculating pOH when given only [H+]
This would require two steps: first take the negative logarithm of the concentration of hydrogen protons. once you get the pH subtract it from 14. this is because PH plus POH equals 14.
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Re: calculating pOH when given only [H+]
If given [H+], use the formula pH=-log[H+] to find pH. Once you find pH, you can subtract it by 14 (14-pH=pOH) to get pOH.
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Re: calculating pOH when given only [H+]
You can find pH from -log[H+] and then subtract 14-pH to find pOH
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Re: calculating pOH when given only [H+]
When only given [H+], you should first find pH with the equation pH = -log[H+], by calculating pH you can then find pOH by subtracting your found pH by 14. pOH = 14 -pH.
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Re: calculating pOH when given only [H+]
First find pH by doing -log([H+]). Then subtract the pH from 14 to get the pOH
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Re: calculating pOH when given only [H+]
Find pH (-log[H+]) first and then subtract pH from 14 for pOH!
Re: calculating pOH when given only [H+]
Since 14 - pH = pOH you can determine the pH by doing -log [H+]. Then subtracting 14 -log [H+] will give you pOH.
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