Page 1 of 1

### pH vs. pOH

Posted: Sun Jan 19, 2020 4:53 pm
What do you do again when you are calculating pOH? Also, how do you know if you are calculating for a base or for an acid?

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 4:56 pm
To calculate pOH, you use the concentration of a base. Alternatively, you calculate the pH using the concentration of an acid. The question may also include a Ka or Kb value. The Ka indicates the equilibrium constant of an acid (uses concentration of hydronium ions), and the Kb indicates the equilibrium constant of a base (uses concentration of hydroxide ions).

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 4:59 pm
The pOH is used when you have the concentration of OH- and is calculated the same way you would calculate pH. You can tell if you are calculating for an acid if it is giving off a proton and you are getting the H3O+ concentration. If it is a base, it would accept a proton and give you the OH- concentration.

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 5:28 pm
to calculate pOH. You find the concentration of OH- molecules and then take the -log[OH-] to find the pOH

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 5:31 pm
You know whether you are calculating for a base or for an acid based on whether there is OH being produced or H3O+.

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 5:33 pm
You calculate pH and pOH by taking the negative log (-log) of [H3O+] or [OH-]. You typically calculate pH for acids and pOH for bases, but questions will often ask you for pH no matter what. Use pH+pOH=14 to convert from pOH to pH.

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 5:50 pm
you use the concentration of a base

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 6:33 pm
to find the pH use the concentration of an acid to find pOH use the concentration of a base. IF given pH you can simply subtract that from 14 to find the pOH

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 6:38 pm
pOH=-log[OH-] and you would use that for bases
pH=-log[H3O+] and you would use this for acids
pH+pOH=14

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 6:39 pm
pH+pOH=14 and pOH = -log[OH-}

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 6:46 pm
Sometimes certain problems will make it easier to calculate pH, and if the pOH is being asked then you can take the pH (-log(H3O+)) and subtract that value from 14 (since pH + pOH + 14).

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 7:23 pm
p before anything means to take the -Log of that thing. pH and pOH are the -log of the H concentrations and the OH concentrations. We can use the K equation to solve for the concentrations. We know we're solving for pH or pOH depending on if the reaction we are using, whether its a weak base or weak acid, and whether its Ka or Kb.

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 7:26 pm
To calculate pOH it is simply =-log[OH-] , and to find pH it is -log[H+] or -log[H30+]. pOH is used to calculate the strength of bases, while pH is used to calculate the strength of acids. To find one if the other is given the equation is pH+pOH=14.

### Re: pH vs. pOH

Posted: Sun Jan 19, 2020 7:35 pm
you find pOH by taking the -log[OH- concentration]. you find pH by taking the -log[H30+ concentration].When it is base you use the OH- concentration and when it is an acid you use the H30+ ion concentration. Also when the equation/problem only gives you one, you can find the other because pH + pOH=14