a) to e) please! Lists of good practice problems from the Harris 8^th ed. text b

Question

a) to e) please! Lists of good practice problems from the Harris 8^th ed. text by chapter, Pdf of problems attached. Reading: all. Exercises: A-D.F.G. Problems: 2,5,8,10,11,18,2026,27,28,32,34,36,40 C_6H_5COOH has a pKa of 4.2. During a summer internship you work in a lab making buffers around pH 4.5. How many grams of sodium benzoate are needed to make one liter of 0.10M stock solution? How many grams of benzoic acid are needed to make one liter of 0.10M stock solution? If you want to make a buffer from your stock solutions with at pH of 4.7, what ratio of acid to base is needed? What volumes of benzoic acid and sodium benzoate stock (parts a & b) are needed to make 250 mL of the buffer in part c? What is the pH of the buffers when the ratio of [A']_eq: [HA]_eq is 1:100, 1:50, 1:10, 1:5, 1:2, 1:1, 2:1, 5:1, 10:1, 50:1, 100:1 ? Plot them.

Explanation / Answer

a)

First number of moles of sodium benzoate:

Moalrity = number of moles / volume

0.10 M = Number of moles / 1.0 L

Number of moles = 0.10*1.0

= 0.10 mol

Amount of sodium bezoate = molar mass * number of moles

=0.10 mol* 144.11 g/mol

= 14.4 g

b)

First number of moles of benzoic acid

Moalrity = number of moles / volume

0.10 M = Number of moles / 1.0 L

Number of moles = 0.10*1.0

= 0.10 mol

Amount of benzoic acid = molar mass * number of moles

=0.10 mol* 122.12 g/mol

= 12.2 g

C)

The pH of the buffer is dependent only on the ratio of the two forms benzoate and benzoic acid which will determine by the use of Henderson-Hasselbalch equation:
pH = pKa + log [benzoate]/[benzoic]

Here pH = 4.7, and pK a= 4.2

pH = pKa + log [benzoate]/[benzoic]

4.7 = 4.2 + log [benzoate]/[benzoic]

log [benzoate]/[benzoic] = 0.5

[benzoate]/[benzoic]=10^0.5

[benzoate]/[benzoic]=3.16

D)

Here [salt]/[acid] = 3.16 or

no. of moles of salt / no. of moles of acid =3.16

molarity = no. of moles of solute / volume of solutionin litres

so no. of moles of solute = molarity X volume of solution in litres

as the buffer solution is of 250ml or 0.250 L

assume that acid be x L ...then salt will be 0.250-x L

no. of moles of acid = 0.250X x
no. of moles of salt = 0.250 X (0.250-x)

put these values in the following equation:

no. of moles of salt / no. of moles of acid = 3.16
0.250(0.250-x) / 0.250x = 3.16

0.0625 -0.250x = 0.79 x

0.0625 = 1.04 x

x = 0.060

x = volume of acid used = 0.060 L or 60 ml
0.250-x = volume of salt used = 0.250-0.060

= 0.19 L or 190 ml

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