sing a primary standard to analyze acid and base solutions

Topics: Hydrochloric acid, Acid dissociation constant, Sodium hydroxide Pages: 7 (1882 words) Published: February 26, 2015

          Using a primary standard to analyze acid and base solutions        Acid-base titration
                 Lab 13G   
Jake Shewchuk              Lab 13C             Dominique Genereux

Purpose 13G:
1. To prepare a standard solution of oxalic acid and use it to standardize an unknown  sodium hydroxide solution. Purpose 13C:
1. To titrate a hydrochloric acid solution of “unknown” concentration with standardized 0.5M sodium hydroxide. 2. To titrate a hydrochloric acid solution of “known” concentration with standardized 0.5M sodium hydroxide. 3. To titrate an acetic acid solution (vinegar) with standardized 0.5M sodium hydroxide. 4. To utilize the titration results to calculate the molarity of the hydrochloric acid and the molarity and percent composition of the vinegar.

Hypothesis 13G: N/A
Hypothesis 13C: N/A
Apparatus & Materials 13G: Please see page 217 in ESSENTIAL EXPERIMENTS for CHEMISTRY Apparatus & Materials 13C: Please see page 196 in ESSENTIAL EXPERIMENTS for CHEMISTRY

Procedure 13G: Please see page 217 in ESSENTIAL EXPERIMENTS for CHEMISTRY Procedure 13C: Please see page 196 in ESSENTIAL EXPERIMENTS for CHEMISTRY

Diagram 13G: N/A
Diagram 13C: N/A

Observation 13G:
Mass of NaOH = 10.836g

Table 1:oxalic acid
Calculated mass of oxalic acid dihydrate
3.150g
Mass of weigh boat (g)
.500g
Mass of weigh boat + oxalic acid (g)
3.662g
Mass of oxalic acid (g)
3.162g
Concentration of 100.00ml
.2510M

Table 2: NaOH
Mass of NaOH + weigh boat
11.336g
Mass of weigh boat
.500g
Calculated mass of NaOH
10.836g
Moles of NaOH
.2709mol
Concentration of 500.0ml NaOH
.5418M

Table 3: HCl
Volume of Concentrated HCl
10.00mL
Concentration diluted to 250.00ml
.5475M

Table 2: Volume of NaOH needed to neutralize 10.00 mL of oxalic acid

Trial 1
Trial 2
Initial reading of buret (mL)
16.39mL
9.89mL
Final reading of buret (mL)
25.98mL
19.50mL
Volume of NaOH required (mL)
9.59mL
9.61mL
Average Volume of NaOH (mL)
9.60mL

Observation 13C:

Table 1: Volume of NaOH needed to neutralize 25.00mL of known HCl solution Molarity of NaOH= .5418M
Trial 1
Trial 2
Initial reading of buret (mL)
9.58mL
3.82mL
Final reading of buret (mL)
34.84mL
29.09mL
Volume of NaOH required (mL)
25.26mL
25.27mL
Average Volume of NaOH (mL)
25.265mL
≃25.26mL

Table 2: Volume of NaOH needed to neutralize 25.00mL of unknown HCl solution Molarity of NaOH= .5418M
Trial 1
Trial 2
Initial reading of buret (mL)
3.94mL
7.92mL
Final reading of buret (mL)
32.79mL
36.78mL
Volume of NaOH required (mL)
28.85mL
28.86mL
Average Volume of NaOH (mL)
28.855mL
≃28.85mL

Table 3: Volume of NaOH needed to neutralize 25.00mL of vinegar Molarity of NaOH= .5418M
Trial 1
Trial 2
Initial reading of buret (mL)
.20mL
.19mL
Final reading of buret (mL)
49.37mL
49.37mL
Volume of NaOH required (mL)
49.17mL
49.18mL
Average Volume of NaOH (mL)
49.175mL
≃49.18mL

Sample Calculations 13G:
Concentration of NaOH=
n=m/MM
n=10.836g/40g/mol
n=.2709mol
C=n/v
C=.2709mol/.5000L
C=.5418M
Sample Calculations 13C:
Concentration of HCl:
C=n/v
C=

Graph 13G: N/A
Graph 13C: N/A

Questions 13G:

Analysis of results 13G:

Part 1: Preparing a primary standard acid solution
1. Calculating one mole of oxalic acid dihydrate
m=n*MM
m=1mol* 126g/mol=126g
2. Calculating the number of moles in the measured mass of oxalic acid.
n=m/MM
n=3.162g/126g/mol
n=.0250952381mol approx=.0251mol
3. Calculating [H2C2O4 • 2H2O] when the mass in procedure part 1, step 1 is dissolved in 250.00mL of solution.
C=n/v
C=.0250952381mol/.1000L
C=.250952381M approx=.2510M

Part 2: Standardizing an unknown NaOH solution
1. Calculate the number of moles of oxalic acid in 25.00mL of standard solution. n=C*v
n=.250952381M*.01000L
n=.0025095238mols   approx=.002510mols
2. Calculating the number of moles of NaOH required to neutralize this amount of oxalic acid...
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