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6.3 – Calculating [H 3 O + ] and [OH - ]

6.3 – Calculating [H 3 O + ] and [OH - ]. Unit 6 – Acids and Bases. Calculating [H + ] and [OH – ].

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6.3 – Calculating [H 3 O + ] and [OH - ]

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  1. 6.3 – Calculating [H3O+] and [OH-] Unit 6 – Acids and Bases

  2. Calculating [H+] and [OH–] • Just as we used Ksp values to help us calculate ion concentrations, we’ll use Ka and Kb 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. • How you calculate ion concentrations depends on whether you have a strong acid (or base) or a weak acid (or base).

  3. 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.

  4. Calculating Ion Concentrations for Strong Acids & Bases • 1. Calculate the hydrogen ion concentration in a 0.050 M solution of hydrochloric acid. • Solution: • We know that HCl is a strong acid that ionizes completely in water (you should memorize the list of strong acids). Begin by writing the balanced reaction: HCl(aq) → H+(aq) + Cl-(aq) • From the balanced equation we see that 1 mole of HCl produces 1 mole of H+ (a 1:1 ratio), therefore the concentration of H+ will equal that of HCl. • Answer: [H+ ] = 0.050 M • Also [Cl-] = 0.050 M

  5. Calculating Ion Concentrations for Strong Acids & Bases • 2. Calculate the hydroxide ion concentration in a 0.010 M solution of barium hydroxide, Ba(OH)2. Barium hydroxide is a strong base. • Solution: • Always begin by writing a balanced equation: Ba(OH)2 (aq) → Ba2+(aq) + 2 OH-(aq) • Since 2 moles of OH- are produced for every 1 mole of Ba(OH)2 , the concentration of OH- will be twice the concentration of Ba(OH)2 . • Answer: [OH-] = 2 × 0.010 = 0.020 M • Also [Ba2+] = 0.010 M

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

  7. Calculating Ion Concentrations for Weak Acids • 3. Calculate the hydrogen ion concentration in a 0.10 M acetic acid solution, HC2H3O2. Ka for acetic acid, a weak acid, is 1.8 ×10-5. • Begin by writing the balanced reaction: HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq) • The question gives us the concentration of the acid, HC2H3O2 (0.10 M). • We need to find the concentration of H+, which will also equal the concentration of C2H3O2- (why?) • Because H+ and C2H3O2- both have a coefficient of “1” • HOWEVER, PLEASE NOTE: ionization is NOT complete because this is a weak acid , [H+] will NOT equal [HC2H3O2]. Instead we must calculate it using the equilibrium constant expression.

  8. Calculating Ion Concentrations for Weak Bases • 4) Calculate the hydroxide ion concentration, [OH-], in a 0.025 M solution of analine, C6H5NH2, a weak base with Kb = 4.3×10-10 • Begin by writing a balanced equation. Since analine is a base that doesn't contain the hydroxide ion, include H2O as a reactant. Also remember that a base is a hydrogen acceptor and will gain an additional H+: 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+].

  9. Equilibrium Constant of Water - Kw • We usually do not think of water as producing ions, but that isn't the case. Water does ionize, although not very well. • We can write the ionization equation for water in two ways (both mean essentially the same thing). The equilibrium constant, Kw, can also be written for both equations as shown. • Notice that H2O does not appear in the Kw expression because it is a liquid, and you'll remember that liquids and solids are not included in equilibrium constant expressions. H2O(l) → H+(aq) + OH-(aq) Kw = [H+] [OH-]=1.0 × 10-14 2 H2O(l) → H3O+(aq) + OH-(aq) Kw = [H3O+] [OH-] =1.0 ×10-14 • *REMEMBER: H+ and H3O+ represent the SAME THING and can be used interchangeably!

  10. Equilibrium Constant of Water - Kw • Two more key items to note: • In pure water, the balanced equation tells us that the concentrations of H+ and OH- will be equal to one another (this is what makes water neutral). Solving the equations shown in the table above, we find that [H+] = [OH-] = 1.0×10-7 • As long as temperature remains constant Kw is a constant - it's value will not change. • The value of Kw is very small, meaning that very few ions are present. • Most water remains "intact" as H2O, and few ions form. Why, then, do we even mention it? • For a very important reason that we will examine in the next section.

  11. Finding [OH-] in Acids and [H+] in Bases • Remember Kw 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 (H+) and some anion (Cl- and C2H3O2- in our previous examples). The concentration of the anion normally doesn't concern us much. • When finding ion concentrations of bases, we determine the concentration of hydroxide ions, OH-, and some cation (Ba2+ and C6H5NH3+ in our base examples above). Again, we aren't too concerned with the cation concentrations.

  12. Finding [OH-] in Acids and [H+] in Bases • Remember water? H2O(l) → H+(aq) + OH-(aq) Kw = [H+] [OH-]=1.0 × 10-14 • 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. • Consider our strong acid example previously, in which [H+] = 0.050 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. • [H+] from water: 1.0×10-7 = 0.000 000 1 • [H+] from HCl + 0.050 • Total [H+] 0.050 0001

  13. Finding [OH-] in Acids and [H+] in Bases • Next, remember that equilibrium constants are A CONSTANT (as long as temperature does not change). Thus the value of Kw will still have a value of 1.0 ×10-14 even though [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= [H+] [OH-] • Rearrange the equation: [OH- ]=Kw • [H+] • Substitute known values and solve for [OH- ]: • [OH- ] =1.0 ×10-14 0.050 • [OH- ] = 2.0 ×10-13 M

  14. 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. H2O(l)↔ H+(aq) + OH-(aq) • When we increase the H+ ion concentration in the water equilibrium system, the reaction will shift to the left to "use up" the additional H+. • This will cause the concentration of OH- to decrease. • Indeed, we see that [OH-] does decrease, from an original concentration of 1.0×10-7 in pure water to 2.0 ×10-13 in our acid solution.

  15. 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 previous slides. • We determined that in a 0.025 M solution of analine, C6H5NH2, the [OH-] was 3.3 ×10-6 M. Our new question - what is the 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 [OH-] and need to find [ H+]. • Again it should be apparent that the contribution to [OH-] from the water, 1.0×10-7 M, MAY have an effect on the total [OH-] always double check:

  16. Use water's equilibrium constant to determine [H+]: • Kw= [H+] [OH-] • Rearrange the equation: • [H+ ] = Kw [OH-] • Substitute in known values and calculate [H+] • [H+] =1.0 ×10-14 3.4 ×10-6 • [H+] =2.9 ×10-9

  17. Facts to Keep in Mind • For any acid or base you can calculate both [H+] and [OH-] • AcidsFirst determine [H+]then use Kwto calculate [OH-] • BasesFirst determine [OH-]then use Kwto calculate [H+] • In water, which is neutral (neither acidic nor basic), [H+] = 1.0×10-7 M and [OH-] = 1.0×10-7 M • Acids increase [H+], so [H+] will be greater than 1.0×10-7 and [OH-] will be less than 1.0×10-7 M • Bases increase [OH-], so [OH-] will be greater than 1.0×10-7, and [H+] will be less than 1.0×10-7 M

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