2.3 Calculating Ion Concentrations - [H+] and [OH–]

We will use K_a and K_b to calculate the concentrations of ions in acids and bases. This will be crucial to determining pH later in this unit, so be sure to follow along carefully.

This is one of the most important sections of this unit mathematically, so be sure you understand every step.

Calculating Ion Concentrations for Strong Acids & Bases

For strong acids and bases, the concentration of the ions can be readily calculated from the balanced equation. Consider these examples carefully.

1.	Calculate the hydrogen ion concentration in a 0.050 M solution of hydrochloric acid.
	Solution: We know that \(\ce{HCl}\) is a strong acid that ionizes completely in water (you should memorize the list of strong acids). Begin by writing the balanced reaction: \(\ce{HCl(aq) -> {H+(aq)} + {Cl-(aq)}}\) From the balanced equation we see that 1 mole of \(\ce{HCl}\) produces 1 mole of\ (\ce{H+}\) (a 1:1 ratio), therefore the concentration of \(\ce{H+}\) will equal that of \(\ce{HCl}\). Answer:[\(\ce{H+}] = 0.050 M\) Also [\(\ce{Cl-}] = 0.050 M\)
2.	Calculate the hydroxide ion concentration in a 0.010 M solution of barium hydroxide, \(\ce{Ba(OH)2}\). Barium hydroxide is a strong base.
	Solution: Always begin by writing a balanced equation: \(\ce{Ba(OH)2(aq) -> {Ba2+(aq)} + {2OH-(aq)}}\) Since 2 moles of \(\ce{OH-}\) are produced for every 1 mole of (\ce{Ba(OH)2}\), the concentration of \(\ce{OH-}\) will be twice the concentration of \(\ce{Ba(OH)2}\) . Answer: [\(\ce{Ba(OH)2}] = 2 \times 0.010 = 0.020 M\) Also: [\(\ce{Ba^{2+}}] = 0.010 M\)

Calculating Ion Concentrations for Weak Acids & Bases

Weak acids and bases require a much different approach to finding ion concentrations. Once you know you have a weak acid or base, follow these steps in finding ion concentrations:

1. Write a balanced equation for the reaction
2. You will need to know the value of K_a or K_b - if it is not given in the question, look it up in a Table of Acid and Base Strengths.
3. Set up the equilibrium constant expression. You will know the value of K_a (or K_b) and the concentration of the acid; you will be solving the equation for the concentration of the ions.

Follow along with these examples very carefully!

1.	Calculate the hydrogen ion concentration in a 0.10 M acetic acid solution, \(\ce{HC2H3O2}\) . Ka for acetic acid, a weak acid, is 1.8 ×10^-5.
	Solution: Begin by writing the balanced reaction: \(\ce{HC2H3O2(aq) <=> {H+(aq)} + {C2H3O2-(aq)}}\) The question gives us the concentration of the acid, \(\ce{HC2H3O2}\) (0.10 M). We need to find the concentration of \(\ce{H+}\), which will also equal the concentration of \(\ce{C2H3O2-}\) (why?) Because ionization is NOT complete because this is a weak acid , [\(\ce{H+}\)] will NOT equal [ \(\ce{HC2H3O2}\)]. Instead we must calculate it using the equilibrium constant expression. Set up the Ka equation: \(Ka = \frac{[\ce{H+}][\ce{C2H3O2-}]}{[\ce{HC2H3O2}]}\) Substitute values into the equation. Let x equal the unknowns \(1.8 \times 10^{-5} = \frac{[x][x]}{[0.10]}\) Rearrange the equation \(x^2 = ({1.8 \times 10^{-5}})({0.10})\) \(x^2 = 1.8 \times 10^{-6}\) Take the square root \(x = 1.3 \times 10^{-3}\) ANSWER \([\ce{H+}] = 1.3 \times 10^{-3} M\) also \([\ce{C2H3O2-}] = 1.3 \times 10^{-3} M\) An important note - in our Ka equation we used a value of 0.10M for [ \(\ce{HC2H3O2}\)]. This isn't quite correct. At equilibrium, the concentration of \(\ce{HC2H3O2}\) will actually be less: initial [ \(\ce{HC2H3O2}\)] = 0.100 subtract the lost [\(\ce{H+}\)] - 0.013 equilibrium [ \(\ce{HC2H3O2}\)] = 0.087 Using this equilibrium concentration of [ \(\ce{HC2H3O2}\)] in our calculations above (instead of 0.10 M), however, does not change our answer significantly. Therefore for our calculations we can safely ignore the negligible change in concentration of the acid.
2.	Calculate the hydroxide ion concentration, [\(\ce{OH-}\)], in a 0.025 M solution of analine, \(\ce{C6H5NH2}\), a weak base with \(Ka = 4.3 \times 10^{-10}\)
	Solution: This question will test some of your skills. Begin by writing a balanced equation. Since analine is a base that doesn't contain the hydroxide ion, include \(\ce{H2O}\) as a reactant. Also remember that a base is a hydrogen acceptor and will gain an additional \(\ce{H+}\): \(\ce{C6H5NH2(aq) + H2O(l) <=> {C6H5NH3+(aq)} + {OH-(aq)}}\) As we did in the previous example, we now set up the Kb expression and solve for ion concentrations. We see from the balanced equation that the ions have a 1:1 ratio, therefore [OH^-] will equal the [C6H5NH3^+]. Set up the Kb equation, omitting liquid water: \(Kb = \frac{[\ce{C6H5NH3+}][\ce{OH-}]}{[\ce{C6H5NH2}]}\) Substitute values into the equation. Let x equal the unknowns \(4.3 \times 10^{-10} = \frac{x}{x}{0.025}\) Rearrange the equation \(x^2 = ({4.3 \times 10^{-10}})({0.25})\) \(x^2 = 1.1 \times 10^{-11}\) Take the square root \(x = 3.3 \times 10^{-6}\) ANSWER \([\ce{OH-}] = 1.3 \times 10^{-3} M\) also \([\ce{C6H5NH3+}] = 1.3 \times 10^{-3} M\)

Finding [OH^-] in Acids and [H⁺] in Bases

Remember K_w from the previous section? Now we learn why it is important.

When we need to determine ion concentrations of an acid, you should immediately realize you will be finding the concentration of hydrogen ion (\(\ce{H+}\)) and some anion (\(\ce{Cl-}\) and \(\ce{C2H3O2-}\) in our acid examples above). The concentration of the anion normally doesn't concern us much.

When finding ion concentrations of bases, we determine the concentration of hydroxide ions, \(\ce{OH-}\), and some cation (\(\ce{Ba^{2+}}\) and \(\ce{C6H5NH3+}\) in our base examples above). Again, we aren't too concerned with the cation concentrations.

Remember water?

\(\ce{H2O(l) -> {H+(aq)} + {OH-(aq)}}\)

\(Kw = [\ce{H+}][\ce{OH-}] = 1.0 \times 10^{-14}\)

Consider our strong acid example from above, in which [H⁺] = 0.05 M. If we add this to the hydrogen ion concentration of pure water, 1.0×10^-7, we get a total hydrogen ion concentration of 0.0500001M. Clearly, the water adds little to the total and we can essentially ignore its contribution.When we make an acid solution, the hydrogen ion concentration will increase because we are adding more H⁺ ions to those already present in water.

[\(\ce{H+}\)] from water:	1.0×10^-7	=	0.000 000 1
[\(\ce{H+}\)] from \(\ce{HCl}\)		+	0.05
Total [\(\ce{H+}\)]			0.050 001

Next, remember that equilibrium constants are A CONSTANT (as long as temperature does not change). Thus the value of K_w will still have a value of 1.0 ×10^-14 even though [\(\ce{H+}\)] has increased due to the presence of the acid. We can use this information to calculate the concentration of hydroxide ions present in the aqueous solution:

	\(Kw = [\ce{H+}][\ce{OH-}]\)
Rearrange the equation	\([\ce{OH-}] = \frac{Kw}{[H+]}\)
Substitute known values and solve for [OH^- ]	\([\ce{OH-}] = \frac{1.0 \times 10^{-14}}{0.05}\)
	\([\ce{OH-}] = 2.0 \times 10^{-13}\)

Le Châtalier's Principle

Remember Le Châtalier's Principle? When we disrupt an equilibrium system by increasing the concentration of a reaction participant, equilibrium will shift to minimize the stress. When we increase the \(\ce{H+}\) ion concentration in the water equilibrium system, the reaction will shift to the left to "use up" the additional \(\ce{H+}\). This will cause the concentration of \(\ce{OH-}\) to decrease. Indeed, we see that [\(\ce{OH-}\)] does decrease, from an original concentration of 1.0×10^-7 in pure water to 2.0 ×10^-13 in our acid solution.

We can apply the same calculations to determine hydrogen ion concentration in any basic solution. Let's return to our weak base, analine, example from above.

We determined that in a 0.025 M solution of analine, \(\ce{C6H5NH2}\), the [\(\ce{OH-}\)] was 1.3 ×10^-3 M. Our new question - what is the \(\ce{H+}\) ion concentration in this basic solution?

To solve, we again refer to our pure water equilibrium and its equilibrium constant expression. This time, however, we have determined [\(\ce{OH-}\)] and need to find [ \(\ce{H+}\)].

Again it should be apparent that the contribution to [\(\ce{OH-}\)] from the water, 1.0×10^-7 M, will have no significant effect on the \(\ce{OH-}\) concentration so we can ignore it from our calculations:

[\(\ce{OH-}\)] from water:	1.0×10^-7	=		0.000 000 1
[\(\ce{OH-}\)] from C6H5NH2	1.3 ×10^-3	=	+	0.001 3
Total [\(\ce{OH-}\)]				0.001 3001

Use water's equilibrium constant to determine [\(\ce{H+}\)]:

	\(Kw = [\ce{H+}][\ce{OH-}]\)
Rearrange the equation	\(\ce{H+} = \frac{Kw}{[OH-]}\)
Substitute in known values and calculate [\(\ce{H+}\)]	\([\ce{H+}] = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-3}}\)
	\([\ce{H+}] = 7.7 \times 10^{-12}\)

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For any acid or base you can calculate both [\(\ce{H+}\)] and [\(\ce{OH-}\)] Acids First determine [\(\ce{H+}\)] then use Kw to calculate [\(\ce{OH-}\)] Bases First determine [\(\ce{OH-}\)] then use Kw to calculate [\(\ce{H+}\)]

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Some things to think about:

• In water, which is neutral (neither acidic nor basic), \([\ce{H+}] = 1.0 \times 10^{-7}\) and \([\ce{OH-}] = 1.0 \times 10^{-7}\)
• Acids increase [\(\ce{H+}\)], so [\(\ce{H+}\)] will be greater than 1.0×10^-7 and [\(\ce{OH-}\)] will be less than 1.0×10^-7M
• Bases increase [\(\ce{OH-}\)], so [\(\ce{OH-}\)] will be greater than 1.0×10^-7, and [\(\ce{H+}\)] will be less than 1.0×10^-7M