pH, pKa, Ka, pKb, Kb

Key Questions

What are pKa and pKb in acids and bases?

$p K_{a}$ $"p"K_"a"$ 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.

Here's a video on $p K_{a}$ $"p"K_"a"$ and $p K_{b}$ $"p"K_"b"$ .

Ernest Z. · 6 · Dec 20 2014
What are Ka and Kb in acids and bases?

Answer:

These are measures of acidity and basicity...

Explanation:

And acid in aqueous solution is conceived to undergo a protonolysis reaction...

$H X (a q) + H_{2} O (l) ⇌ H_{3} O^{+} + X^{-}$ $HX(aq) + H_2O(l) rightleftharpoonsH_3O^+ + X^(-)$

And as for any equilibrium, we can measure and quantify it in the usual way...

$K_{a} = \frac{[H_{3} O^{+}] [X^{-}]}{[H X (a q)]}$ $K_a=([H_3O^+][X^-])/([HX(aq)])$

Note that $H_{2} O$ $H_2O$ DOES NOT appear in the equilibrium expression because it is present in such high concentration that it is effectively constant..

For strong acids, i.e. $H I$ $HI$ , $H B r$ $HBr$ , $H C l$ $HCl$ , $H_{2} S O_{4}$ $H_2SO_4$ protonolysis is effectively quantitative: the given equilibrium lies entirely to the right as we face the page, and the acid solution is quantitative in $H_{3} O^{+}$ $H_3O^+$ . For weaker acids, $H F$ $HF$ , $H_{3} C - C O_{2} H$ $H_3C-CO_2H$ , the equilibrium lies somewhat to the left...and concentrations of the parent acid remain at equilibrium.

And likewise, we can formalize the performance of a base by an equivalent equilibrium...we use ammonia, because this is a WEAK base in aqueous solution...

$N H_{3} (a q) + H_{2} O (l) ⇌ N H_{4}^{+} + H O^{-}$ $NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + HO^-$

And $K_{b}$ $K_b$ is defined in an equivalent way to $K_{a}$ $K_a$ ...

$K_{b} = \frac{[N H_{4}^{+}] [H O^{-}]}{[N H_{3} (a q)]}$ $K_b=([NH_4^+][HO^-])/([NH_3(aq)])$ , $K_{b} (ammonia) = 1.74 \times 10^{- 5}$ $K_b"(ammonia)"=1.74xx10^-5$ ...

Confused yet....?

Well, note that NECESSARILY....for a given acid/conjugate pair, say $N H_{4}^{+} / N H_{3}$ $NH_4^+"/"NH_3$ ...

$K_{a} K_{b} = 10^{- 14}$ $K_aK_b=10^-14$ ...or perhaps more usefully...

$p K_{a} + p K_{b} = 14$ $pK_a+pK_b=14$ ...

anor277 · 1 · Jun 18 2018
What is pH in acids and bases?

The pH scale provides a way of measuring how acidic or basic solutions are. The scale ranges from 0-14. A pH of 0 is the most acidic, 7 is neutral and 14 is the most basic.

Here is a video of a lab which looks at a number of different solutions and measures their pH levels using a pH meter and an indicator.

video from: Noel Pauller

mrpauller.weebly.com · 2 · Jan 9 2015

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