TECHNIQUES AND CORRELATIONS LEARNING
WORKBOOK 2026 ANALYTICAL METHODS
AND BIOCHEMICAL TESTING
◉ What is the normality for a solution containing 100 g of NaCl
made up to 500 mL with distilled water? Assume a gram molecular
weight (from periodic table) of approximately 58 grams.
a. 3.45
b. 0.86
c. 1.72
d. 6.9. Answer: a. 3.45
◉ What is the percent (w/v) for a solution containing 100 g of NaCl
made up to 500 mL with distilled water?
a. 20%
b. 5%
c. 29%
d. 58%. Answer: a. 20%
◉ What is the dilution factor for a solution containing 100 g of NaCl
made up to 500 mL with distilled water?
,a. 1:5 or 1/5
b. 5
c. 50 or 1/50
d. 10. Answer: a. 1:5 or 1/5
◉ What is the value in mg/dL for a solution containing 10 mg of
CaCl2 made with 100 mL of distilled water?
a. 10
b. 100
c. 50
d. Cannot determine without additional information. Answer: a. 10
◉ What is the molarity of a solution containing 10
mg of CaCl2 made with 100 mL of distilled water? Assume a gram
molecular weight from the periodic table of approximately 111
grams.
a. 9 × 10-4
b. 1.1 × 10-3
c. 11.1
d. 90. Answer: a. 9 x 10-4
◉ You must make 1L of 0.2M acetic acid(CH3COOH). All you have
available is concentrated glacial acetic acid (assay value, 98%;
,specific gravity, 1.05 g/mL). It will take ______ milliliters of acetic acid
to make this solution. Assume a gram molecular weight of 60.05
grams.
a. 11.7
b. 1.029
c. 3.42
d. 12.01. Answer: a. 11.7
◉ What is the hydrogen ion concentration of an acetate buffer
having a pH of 4.24?
a. 5.75 × 10-5
b. 1.19 × 10-1
c. 0.62
d. 0.76 × 10-4. Answer: a. 5.75 x 10-5
◉ Using the Henderson-Hasselbalch equation, give the ratio of salt
to weak acid for a Veronal buffer withapHof8.6andapKa of 7.43.
a. 14.7/1
b. 1/8.6
c. 1.17/1
d. 1/4.3. Answer: a. 14.7/1
, ◉ The pKa for acetic acid is 4.76. If the concentra- tion of salt is 5
mmol/L and that of acetic acid is 10 mmol/L, what is the expected
pH?
a. 4.46
b. 5.06
c. 104
d. 56. Answer: a. 4.46
◉ The hydrogen ion concentration of a solution is 0.0000937. What
is the pH?
a. 4.03
b. 9.37 × 10-5
c. 9.07
d. 8.03. Answer: a. 4.03.
◉ Perform the following conversions:
4×104 mg=___g
1.3×102 mL=____dL
0.02 mL = ____ μL
5×10-3 mL=___μL
5×10-2 L=___mL
4 cm = _____ mm. Answer: 4×104 mg= 40 g