Chem 220 - Prelab Exp 2A – Due February 6th by midnight Perform the following ca

Question

Chem 220 - Prelab Exp 2A – Due February 6th by midnight

Perform the following calculations before arriving in class (in Excel or on paper/by hand on separate notebook paper). Upload all your calculations in ELN in Experiment #2A prelab 1. You must show your work for full credit.

Label each of the following compounds as either strong acid (sa) , weak acid (wa), strong base (sb), weak base (wb), amphoteric (pp) or neutral salt (n)

(a) 0.05 M NaCl

(b) 0.05 M Acetic Acid

(c) 0.05 M Sodium Acetate

(d) 0.05 M KH2PO4

(e) 0.05 M Tris [(CH2OH)3CNH2]

(f) 0.05 M Na2CO3

(g) 0.05 M HCl

Calculate the original pH of 25 mL of each solution

Calculate the pH following the addition of 1 mL of 0.10 M HCl to 25 mL of each solution

Calculate the pH following the addition of 1 mL of 0.10 M NaOH to 25 mL of each solution.

Create a summary table.

Example:

Table 1: Calculated values of pH for simple solutions (Submitted in prelab)

Sol.

Name

Species Type*

Original pH of 25 mL of solution

pH of 25 mL after adding 1.0 mL of
0.1M HCl

pH of 25 mL after adding 1.0 mL of
0.1 M NaOH

a

0.05 M NaCl

b

0.05 M HC2H3O2

c

0.05 M NaC2H3O2

d

0.05 M KH2PO4

e

0.05 M Tris

f

0.05 M Na2CO3

g

0.05 M HCl

* strong acid (sa) , weak acid (wa), strong base (sb), weak base (wb), amphoteric (pp) or neutral salt (n)

Sol.

Name

Species Type*

Original pH of 25 mL of solution

pH of 25 mL after adding 1.0 mL of
0.1M HCl

pH of 25 mL after adding 1.0 mL of
0.1 M NaOH

a

0.05 M NaCl

b

0.05 M HC2H3O2

c

0.05 M NaC2H3O2

d

0.05 M KH2PO4

e

0.05 M Tris

f

0.05 M Na2CO3

g

0.05 M HCl

Explanation / Answer

For the given solutions

a. 0.05 M NaCl

Neutral salt

initial pH = 7

pH does not change upon addiiton of HCl or NaOH

b. 0.05 M HC2H3O2

weak acid

initial pH

[H+] = sq.rt.(1.8 x 10^-5 x 0.05) = 9.5 x 10^-4 M

pH = -log[H+] =  3.02

after adding moles of HCl = 0.1 M x 1 ml = 0.1 mmol to HC2H3O2 = 0.05 M x 25 ml = 1.25 mmol

pH = -log[H+] = -log(0.1) = 1

after adding 1 ml of 0.1M NaOH

[HC2H3O2] remining = 1.24 mmol/26 ml = 0.05 M

[NaC2H3O2] formed = 0.1 mmol/26 ml = 0.004 M

pH = pKa + log(base/acid)

      = 4.75 + log(0.004/0.05) = 3.64

c. 0.05 M NaC2H3O2

strong base

initial pH

[OH-] = sq.rt.(5.55 x 10^-10 x 0.05) = 5.3 x 10^-6 M

pOH = -log[OH-] = 5.3

pH = 14 - pOH = 8.72

after adding 0.1 mmol HCl

[NaC2H3O2] remining = 1.24 mmol/26 ml = 0.05 M

[HC2H3O2] formed = 0.1 mmol/26 ml = 0.004 M

pH = pKa + log(base/acid)

      = 4.75 + log(0.05/0.004) = 5.85

after adding 1 ml of 0.1 M NaOH

pOH = -log[OH-] = 1

pH = 14 - 1 = 13

Similarly other solutions can be done

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