Acid-Base Chemistry

Aim Prepare standard NaOH and HCl to detect the concentration of acetic astringent in vinegar as well as ammonium hydroxide water in window cleaner via titration using sufficient indicators. Results and Calculations 1.1. Titration condition 1: normalisation of NaOH 1.2.1 Results of Titration of KHC8H4O4 against NaOH Titration| 1| 2| 3| last record handwriting of Titrant/mL| 8.60| 13.55| 18.50| Initial stack of Titrant/mL| 3.20| 8.60| 13.55| al-Quran of Titrant physical exercise/mL| 5.40| 4.95| 4.95| | | | | Indicator use| Phenolphthalein| colouring material smorgasbord at resultant| Colourless to pinko| 1.2.2 Calculations Mass of KHC8H4O4 utilize= 1.0291 g poor boy Mass of KHC8H4O4= 204.22 g/mol yard 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 just pot of NaOH titrant apply = 4.95 mL chemical compare of response: KHC8H4O4 (aq) + NaOH (aq) KHC8H4O3-Na+ (aq) + water (l) KHC8H4O4 reacts with NaOH in a 1:1 ratio, thus nary(prenominal) 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 narrow 2: Standardization of HCl 1.3.3 Results of Titration of HCl against NaOH Titration| 1| 2| Final record 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 out 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 stay put a full essay, pasture it on our website: Ordercustompaper.com

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