Question

1. Calculate [OH−] for a solution formed by adding 5.00 mL of 0.120 M KOH to 20.0 mL of 7.1×10⁻² M Ca(OH)2.

2. Calculate pH for a solution formed by adding 5.00 mL of 0.120 M KOH to 20.0 mL of 7.1×10⁻² M Ca(OH)2.

3. Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows:

C5H5N+H2O⇌C5H5NH++OH−

The pKb of pyridine is 8.75. What is the pH of a 0.430 Msolution of pyridine?

4. Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid dissociates in water:

C6H5COOH⇌C6H5COO−+H+

The pKa of this reaction is 4.2. In a 0.79 M solution of benzoic acid, what percentage of the molecules are ionized?

Answer 1

1.

KOH releases 1 OH, so you are adding:

0.005Lx0.120mol/L=6x10^-4molof OH-

Ca(OH)2 releases 2 OH per mole, so you are adding:

2x0.020Lx7.1×10⁻²mol/L=2.84x10^-3mol

In the final solution you 2.84x10^-3+6x10^-4=3.44x10^-4mol OH

divided by the final volume

2)

pOH=-log[OH^-]=-log[0.1376]=0.8614

pH=14-pOH=14-0.8614=13.1386

3)

from the reaction and the mass action equation:

you should analyze the intial and equilibrium concentrations:

	C5H5N	C5H5NH+	OH−
intitial	0.430	0	0
equilibrium	0.430-x	x	x

x is the concentration formed of C5H5NH+ and OH−, thus the concentration lost by the pyridine

and using the mass action equation

pKb=-logKb----->Kb=10^(-pKb)

<-----this yields a second degree equation:

<-----from the equilibrium analysis

<----again as above

4)

An analysis similar to number 3 but with acid dissociation:

pKa=-logKa----->Ka=10^(-pKa)

	C6H5COOH	C6H5COO−	H+
intitial	0.79	0	0
equilibrium	0.790-x	x	x