ABSE_pH of Dilute Strong Acids Lesson

pH of Dilute Strong Acids

If you do the calculation to find -log (1.5 x 10-10), you get an answer of 9.8. Always stop to ask yourself if an answer makes sense. Does it?  It doesn't!

The solution is made only of water and HBr, an acid. So, it is not a basic solution even though the math showed a pH greater than 7. We must be missing something. What we are missing here is the contribution of H+ ions from the auto-ionization of water. In an acidic or basic solution, there are two sources of H+ ions.

[H+]Total = [H+]from solute   +   [H+]from water

In the calculations on acid or base solutions on the previous page, we neglected any contribution of either H+ or OH- ions to the overall solution because those values are not significant. We assumed the equation to be:

 [H+]Total = [H+]from solute +  zero from water

 In the "Think About It" example above, the concentration of the solution is so small that the contribution of H+ ions from the auto-ionization of water is significant. So, we must add both sources of H+ to get the total and calculate the pH.

[H+]Total = [H+] from solute + [H+] from water [H+]Total = 1.5 x 10-10 + 1.0 x 10-7 [H+]Total = 1.0015 x 10-7 pH = 6.999
Using significant digits, we would round this to 7.0. More digits were left here just to show how that the solution is in fact acidic.

We only need to approach a pH (or pOH) calculation in the manner shown above when we are dealing with very dilute solutions of strong acids or bases.  Recall the list of strong acids and strong bases that you learned back in module 3 on chemical reactions.

MEMORIZE
Strong acids
HCI - hydrochloric acid
HBr - hydrobromic acid
HI - hydroiodic acid
HNO3 nitric acid
H2SO4 - sulfuric acid
HCIO4 - perchloric acid
HCIO3 - chloric acid
Strong bases
Hydroxides of group IA and IIA

Calculating pH or pOH of these strong acids and strong bases are normally straight forward. We use these equations as needed:

pH = -log[H+]

pOH = -log[OH-]

pkw = -log kw

pH + pOH = pkw

pH + pOH = 14

kw = [H3O+][OH-] = 1.0 x 10-14

Remember as we calculate pH, that the terms strong or weak do not refer to concentration, rather they refer to complete dissociation in water.  

Weak Acids and Bases

While strong acids and bases ionize completely, weak ones do not. Below is an illustration of this. Notice that the strong acid forms many ions and the weak one does not.

Weak acid, only a small proportion of the molecules give up their protons
Strong acid, most of the molecules give up their protons
diagrams of weak and strong acids

 

Weak acids and bases therefore exist in equilibrium with the ions formed by their reaction with water.

The general ionization equation of a weak acid in water is:

HA   +   H2O ⇌  H3O+   +   A-

The equilibrium expression for this reaction is:

LaTeX: k_{eq}=\frac{[H_3O^+][A^-]}{[H_2O][HA]}keq=[H3O+][A][H2O][HA]

Moving [H2O] to the side with keq gives us:

LaTeX: k_a=\frac{[H_3O^+][A^-]}{[HA]}ka=[H3O+][A][HA]

You must be able to write the ka equation and equilibrium expression for any weak acid.  

Example Write the ionization equation for acetic acid, HC2H3O2(aq) and the ka expression.
First, write the weak acid plus water.
HC₂H₃O₂ + H₂O
Remember that the acid donates an H+. Use this concept to determine the products.
HC₂H₃O₂ + H₂O H₃O + C₂H₃O₂⁻
ka [H₂O⁺][C₂H₂O₂⁺]/[HC₂H₃O₂]
With weak acids, you can write an abbreviated form of the equation and the ka expression by leaving out the water in the equation.
HC₂H₃O₂ = H+ + C₂H₃O₂⁻
ka [H*][C₂H₃O₂⁺]/[HC₂H₃O₂]

The general ionization equation of a weak base in water is:

B   +   H2O ⇌  BH+   +   OH-

The equilibrium expression for this reaction is:LaTeX: k_{eq}=\frac{[BH^+][OH^-]}{[H_2O][B]}keq=[BH+][OH][H2O][B]

Moving [H2O] to the side with keq gives us:LaTeX: k_{eq}=\frac{[BH^+][OH^-]}{[B]}keq=[BH+][OH][B]

You must be able to write the kb equation and equilibrium expression for any weak base.

Example
Write the ionization equation for ammonia, NH₃(aq) and the kь expression.
First, write the weak base plus water.
NH₃ + H₂O
Remember that the base accepts an H+. Use this concept to determine the products.
NH₃ + H₂O↔️ NH₄⁺ + OH-
k_b= [NH₄⁺][OH-]/[NH₃]
With weak bases, you can't write an abbreviated form of the equation. You must add the water!

pka and pkb

The value of ka and kb are usually small numbers and can be expressed in log form (like pH). pka = -log ka

Here are some common acids with their ka and pka values.

ka and pK_a Values for Weak Monoprotic acids at 25°C

The larger the kx, the stronger the acid or base.

The larger the pkx, the weaker the acid or base.

There is an interesting relationship between the equilibrium constants for a conjugate acid-base pair. Let's write the equilibrium expressions for HCHO2 and CHO2- to see this relationship.

Write the ka and kb for HCHO2 and CHO₂ to determine ka x kb.
HCHO₂ + H₂O CHO₂ + H₃O+
k_a = [H₃O⁺][CHO₂ ]/[HCHO₂]
CHO₂⁻ + H₂O ↔️ HCHO₂+ OH-
kb= [HCHO₂][OH⁻]/[CHO₂]
ka x ko = [H₃O⁺][CHO₂⁺] x [HCHO,][OH] X
[HCHO,] [CHO,]
Ka X Kb = [H₃O⁺][CHO, ]/[HCHO₂]X[HCHO₂] [OH⁻]/[CHO,]
Ka X kb = [H₂O⁺] x [OH⁻]
This is the same as kw.
This relationship exists for any acid-base conjugate pair.

Because of this relationship, the following is also true:

pka   +   pkb   =   pkw   = 14

pka   = -log ka

pkb   = -log kb

Ka X Kb = kw
pka + pkb = pkw pkw = 14
pka = -log ka
pkb = -log Kb

 

Remember to work on the module practice problems as you complete each section of content.  

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