Acid-Base Chemistry

Aim
Prepare standard NaOH and HCl to determine the concentration of acetic sulfurous in vinegar as well as ammonia water in window cleaner via titration using suitable indicators.

Results and Calculations
1.1. Titration condition 1: Standardization of NaOH
1.2.1 Results of Titration of KHC8H4O4 against NaOH
Titration| 1| 2| 3|
last record book of Titrant/mL| 8.60| 13.55| 18.50|
Initial Volume of Titrant/mL| 3.20| 8.60| 13.55|
Volume of Titrant use/mL| 5.40| 4.95| 4.95|
| | | |
Indicator used| Phenolphthalein|
Colour smorgasbord at Endpoint| Colourless to Pink|
1.2.2 Calculations
Mass of KHC8H4O4 used= 1.0291 g
Molar Mass of KHC8H4O4= 204.22 g/mol
Molarity of KHC8H4O4= (1.0291 ÷ 204.22) ÷ 0.1000
= 0.05039 M
zero(prenominal) of moles of KHC8H4O4 in 10.0 mL= (10.0 ÷ 1000) × 0.05039
= 5.039 ×10-4 mol

Average pot of NaOH titrant used = 4.95 mL

Chemical compare of response: KHC8H4O4 (aq) + NaOH (aq) KHC8H4O3-Na+ (aq) + H2O (l)
KHC8H4O4 reacts with NaOH in a 1:1 ratio, thus
No. of moles of NaOH in 4.95 mL= 5.039 ×10-4 mol
Concentration of NaOH= (5.039 ×10-4) ÷ (4.95÷ 1000)
= 0.1018 M

1.2.



Titration Set 2: Standardization of HCl
1.3.3 Results of Titration of HCl against NaOH
Titration| 1| 2|
Final Volume of Titrant/mL| 29.20| 39.95|
Initial Volume of Titrant/mL| 18.50| 29.25|
Volume of Titrant used/mL| 10.70| 10.70|
| | |
Indicator used| Methyl Orange|
Colour transfer at Endpoint| Red to Yellow|

1.3.4 Calculations
Average volume of NaOH titrant used= 10.70 mL
No. of moles of NaOH used= (10.70 ÷ 1000) × 0.1018
= 1.089 ×10-3 mol

Chemical Equation of Reaction: HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
HCl reacts with NaOH in a 1:1 ratio, thus
No. of moles of HCl in 10.0 mL = 1.089 ×10-3 mol
Concentration of HCl= (1.089 ×10-3) × (1000÷...If you want to get a full essay, order it on our website: Orderessay



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