Use formulas if you need to, but words are preferable. I don't need to know how they're used in formulas as much as what they represent. Like, I know that pH is the acidity (low pH is high acidity), but how does that relate to pOH, pKa, and Ka? But not just those, how they all relate. I don't really understand that. :/
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Hydrogen ion concentration in moles/liter is [H+]. Hydroxide ion concentration in moles per liter is [OH-].
pH is the negative logarithm of [H+], -Log[H+]. pOH is -Log[OH-].
Pure water H2O ionizes very slightly to H+ and OH-. In pure water, [H+] = 1 x10^-7, and [OH-] = 1 x 10^-7. Log(1x10^-7) = -7, so -Log[H+] in pure water is 7. That is why the pH of a neutral solution is 7.
The equilibrium constant for ionization of pure water Kw is [H+][OH-]. (1x10^-7) x (1x10^-7) = 1 x 10^-14, so Kw = 1 x 10^-14.
Adding the logs of two numbers together gives you the product of the two numbers, so pH + pOH = 14.
If you have a weak acid like acetic acid, HC2H3O2, then it ionizes slightly according to HC2H3O2 ===> H+ + C2H3O2-. The Ka for acetic acid is [H+][C2H3O2-]/[HC2H3O2] = 1.8 x 10^-5.
pKa is -LogKa. If you enter 1.8 x 10^-5 into a calculator and press, you get -4.7. So pKa is +4.7 for acetic acid to two significant figures.
pH is the negative logarithm of [H+], -Log[H+]. pOH is -Log[OH-].
Pure water H2O ionizes very slightly to H+ and OH-. In pure water, [H+] = 1 x10^-7, and [OH-] = 1 x 10^-7. Log(1x10^-7) = -7, so -Log[H+] in pure water is 7. That is why the pH of a neutral solution is 7.
The equilibrium constant for ionization of pure water Kw is [H+][OH-]. (1x10^-7) x (1x10^-7) = 1 x 10^-14, so Kw = 1 x 10^-14.
Adding the logs of two numbers together gives you the product of the two numbers, so pH + pOH = 14.
If you have a weak acid like acetic acid, HC2H3O2, then it ionizes slightly according to HC2H3O2 ===> H+ + C2H3O2-. The Ka for acetic acid is [H+][C2H3O2-]/[HC2H3O2] = 1.8 x 10^-5.
pKa is -LogKa. If you enter 1.8 x 10^-5 into a calculator and press
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p slept with H and OH , Whilst being married to Ka.