Chapter 17

1. Chapter 17 Part 2

2. Salt Solutions We will look at the cation and the anion separately, and then combine the result to determine whether the solution is acidic, basic, or neutral.

3. The conjugate acid of a strong base is very weak and does not react with water. It is therefore considered to be neutral. Na+ + H2O  No Reaction (NR) The conjugate base of a strong acid is very weak and does not react with water. It is therefore considered to be neutral. Cl- + H2O  NR

4. The conjugate acid of a weak base reacts with water to form an acidic solution: NH4+ + H2O <==> NH3 + H3O+ The conjugate base of a weak acid reacts with water to form a basic solution: F- + H2O <==> HF + OH-

5. We classify each salt by examining its cation and its anion, and then combining the result. NaBr Na+ is the conjugate acid of NaOH, a strong base. It does not react with water, so it is --------- Br- is the conjugate base of HBr, a strong acid. It does not react with water, so is ----------- The cation is neutral; the anion is neutral. NaBr is ---------------

6. NaC2H3O2 Na+ is the conjugate acid of NaOH, a strong base. It does not react with water, so it is _________ C2H3O2- is the conjugate base of HC2H3O2-, a weak acid. It reacts with water to give a ________ solution. The cation is neutral; the anion is basic. NaC2H3O2 is _________

7. NH4Cl NH4+ is the conjugate acid of NH3, a weak base. It reacts with water to give an ________ solution. Cl- is the conjugate base of HCl, a strong acid. It does not react with water, so it is _________ The cation is acidic; the anion is neutral. NH4Cl is __________

8. NH4F NH4+ is the conjugate acid of NH3, a weak base. It reacts with water to give an ________ solution. F- is the conjugate base of HF, a weak acid. It does not react with water, so it is ________ The cation is acidic; the anion is basic. We need more information in this case. We compare Ka for HF to Kb for NH3.

9. If Ka > Kb, the solution is acidic. If Ka < Kb, the solution is basic. If Ka = Kb, the solution is neutral.

10. Ammonium nitrate, NH4NO3, is administered as an intravenous solution to patients whose blood pH has deviated from the normal value of 7.40. Would this substance be used for acidosis (blood pH < 7.40) or alkalosis (blood pH > 7.40)? NH4+ is the conjugate acid of a NH3, a weak base. NH4+ is _______ NO3- is the conjugate base of HNO3, a strong acid. NO3- is _______ NH4NO3 is acidic, so it could be used for ___________

11. The hydrolysis equilibrium constant can be used in problems to determine the pH of a salt solution. To use the hydrolysis equilibrium, we need to compute the K value for it.

12. What is Kb for the F- ion, the ion added to the public water supply to protect teeth? For HF, Ka = 6.8 × 10-4.

13. Household bleach is a 5% solution of sodium hypochlorite, NaClO. This corresponds to a molar concentration of about 0.70 M NaClO. What is the [OH-] and the pH of the solution? For HClO, Ka = 3.5 × 10-8.

14. We are told that Ka = 3.5 × 10-8. That means that Kb= 2.9 × 10-7. This allows us to substitute into the Kb expression to solve for x.

16. Common-Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that takes part in the equilibrium. • Buffers • A buffer solution is characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. • A buffer is made by combining a weak acid with its conjugate base or a weak base with its conjugate acid.

17. What is the pH of a buffer made by mixing 1.00 L of 0.020 M benzoic acid, HC7H5O2, with 3.00 L of 0.060 M sodium benzoate, NaC7H5O2? Ka for benzoic acid is 6.3 × 10-5. This problem involves dilution first. Once we know the concentrations of benzoic acid and benzoate ion, we can use the acid equilibrium to solve for x. We will use HBz to represent benzoic acid and Bz- to represent benzoate ion.

19. Calculate the pH of a 0.10 M HF solution to which sufficient sodium fluoride has been added to make the concentration 0.20 M NaF. Ka for HF is 6.8 × 10-4. We will use the acid equilibrium for HF. NaF provides the conjugate base, F-, so [F-] = 0.20 M.

21. Henderson–Hasselbalch Equation Buffers at a specific pH can be prepared using the Henderson-Hasselbalch Equation. Buffers are prepared from a conjugate acid-base pair in which the ionization is approximately equal to the desired H3O+ concentration. Consider a buffer made up of a weak acid HA and its conjugate base A- The acid-ionization constant is

22. The preceding equation can be used to derive an equation for the pH of the buffer. Take the negative logarithm of both sides of the equation. The pKa of a weak acid is defined in a manner similar to pH and pOH pKa = - log Ka

23. The equation can then be expressed as: This equation is generally shown as: This equation relates buffer pH for different concentrations of conjugate acid and base and is known as the Henderson-Hasselbalch equation.

24. Question: What is the [H3O+] for a buffer solution that is 0.250 M in acid and 0.600 M in the corresponding salt if the weak acid Ka = 5.80 x 10-7? Use as example the following equation of a conjugate acid-base pair

25. Acid–Base Titration An acid–base titration is a procedure for determining the amount of acid (or base) in a solution by determining the volume of base (or acid) of known concentration that will completely react with it.

26. An acid–base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). Such curves are used to gain insight into the titration process. You can use the titration curve to choose an indicator that will show when the titration is complete.

27. This titration plot shows the titration of a strong acid with a strong base. Note that the pH at the equivalence point is 7.0.

28. The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added. The indicator must change color near the pH at the equivalence point.