What are pKa and pKb in acids and bases?

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Answer 1 · 2014-12-20T19:46:09

$p K_{a}$ $"p"K_"a"$ and $p K_{b}$ $"p"K_"b"$ are measures of the strengths of acids and bases, respectively

Acids

When you dissolve an acid in water, it undergoes an equilibrium reaction with the water in an.

HA + H₂O ⇌ H₃O⁺ + A⁻

The value of the equilibrium constant is given by

$K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{HA}$ $K_"a" = (["H"_3"O"^+]["A"^-]]/["HA"]$

The greater the value of $K_{a}$ $K_"a"$ , the stronger the acid.

For most weak acids, $K_{a}$ $K_"a"$ ranges from $10^{- 2}$ $10^-2$ to $10^{- 14}$ $10^-14$ .

We convert these exponential numbers into a normal range by taking their negative logarithm.

The operator $p$ $"p"$ means "take the negative logarithm of".

So $p K_{a} = - {log K}_{a}$ $"p"K_"a" = -logK_"a"$ .

For most weak acids, $p K_{a}$ $"p"K_"a"$ ranges from 2 to 14.

Thus, the smaller the value of $p K_{a}$ $"p"K_"a"$ , the stronger the acid.

Bases

When you dissolve a base in water, it reacts with the water in an equilibrium reaction.

B + H₂O ⇌ BH⁺ + OH⁻

The value of the equilibrium constant is given by

$K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{B}$ $K_"b" = (["BH"^+]["OH"^-]]/["B"]$

The greater the value of $K_{b}$ $K_"b"$ , the stronger the base.

For most weak acids, $K_{b}$ $K_"b"$ ranges from $10^{- 2}$ $10^-2$ to $10^{- 13}$ $10^-13$ .

$p K_{b} = - {log K}_{b}$ $"p"K_"b" = -logK_"b"$ .

For most weak acids, $p K_{a}$ $"p"K_"a"$ ranges from 2 to 13.

The smaller the value of $p K_{b}$ $"p"K_"b"$ , the stronger the base.

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