K to pH

[Andrew Nguyen 2K]

Does anyone have a good way to remember how to find pH given Ka, Kb, and Kc?

[Allison Peng 1D]

I think that in order to find pH if you're given Ka, for a monoprotic acid Ka=[hydronium]*[conjugate base]/[acid] at equilibrium, so you can use pH to find [hydronium] and then use ice tables and the given Ka value to find the concentrations of acid and conjugate base at equilibrium. or, if you're given initial [acid], you can also use ice tables and Ka to find pH.

If the substance in question is a base and you're given Kb, for a monoprotic base Kb=[hydroxide]*[conjugate acid]/[base] @ eq, you can use pH (converted to [hydroxide]) and Kb to find concentrations of base and conjugate acid at equilibrium. (Can find pH from Ka and initial concentrations in the same way as above).

I think Kc is just the equilibrium constant for some reaction, so this can either be Ka or Kb.

Additionally, sometimes you are given Ka or Kb of the conjugate. In this case, convert it to its complement and use that Kb or Ka value to find equilibrium concentrations.

[Joseph Lee]

-log[Ka] = pH

-log[Kb] = pOH

-log[Kc] = pH or pOH (depends on whether it is a weak acid or base reaction

[Celine Khuu 2F]

When you're given Ka/Kb and an initial concentration of an acid/base, the first thing you want to do is write down a chemical equation that depicts what is going on in the reaction. For example, HA -> H + A would represent the dissociation of a weak acid. Then, you would proceed with your ice box and plug in the values you get from the "e" portion into your equilibrium constant expression. From this, you should solve for the value of x using the Ka/Kb you were given. Typically, x would equal the concentration of [OH] or [H3O] of the mixture at equilibrium and you would take the -log of that value to get your pH or pOH.

[605721741]

Ka = -log(H+)
Kb = Kw/Ka
Kw = 1 * 10^-14

[405532435]

To find the pH, you can take the -log of the value Ka, Kb, or Kc.
Hope this helps!

[Jordan Kaplan 2E]

You can't just take the -log(Ka,Kb, or Kc) to find the pH or pOH since the pH is the [H3O+] and pOH is [OH-]. If you were given Kb or Ka, you would need to set up an ice table and solve for the concentration of H3O+ or OH-, depending on whether the reaction is acidic or basic.

Taking the -log(Ka) will give you the pKa, and same for Kb. the "p" just stands for -log, but you cannot just go straight from Ka to pH!

[105778693]

For Ka, use -log [Ka] = pH.
For Kb, you would find pOH (-log [Kb] = pOH) before finding pH (pH + pOH = 14).
For Kc, use -log [Kc] = pH or pOH.

[305597516]

Since the Ka refers to acids, you can use it to find the concentration of H3O+. You can then use the equation -log(H3O+) to find the pH.

The Kb refers to bases, so you can use this to find the concentration of OH-. By using the -log function, you can find the pOH. Since pH + pOH = 14, you can subtract the pOH from 14 to find the pH.

Ka and Kb are also related in that Ka x Kb = Kw, so if you have one, you can divide it from 1.0 x 10^-14 to find the other, which should allow you to find the pH through the suggestions above. Hope this helped!

[405825570]

pH=-log[H+]
pOH=-log[OH-]
Kc=1x10-14=KaKb

[sidneypalacios]

If you have kA then just take the negative log to find pH. If you have Kb, you would take the negative log of Kb to find pOH, then use this equation (14=pH+pOH) to find pH. Lastly for Kc you would just take the negative log of Kc to find either pH or pOH depending on whether it is a weak acid or base