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• Acid-base titration is a process for calculating the concen tration of a known volume of acid or base. H ( 4 0 C 2 aq) acid + 2 NaOH(aq) —> 2 N ( 4 0 aq aC ) + 22 8 0 (Iiq) base Carry out this reaction using a TITRATION. Setup for titrating an acid with a base Sample Prob lem • In this sample titration , we determine the concentra are trying to HCI. In the titration we wil tion of 20.00 mL of HCI with 0.150 M NaOH. l be neutralizing the • Step 1: The NaOH, the buret. The titrant is the titrant, is placed in the concentration that is addsolution of known ed from the buret. • Step 2: The HCI is plac ed in the Erlenmeyer flask along with approxima distilled water and 2-3 dro tely 20.00 mL of ps of phenolphthalein indicator. Since the solution phenolphthalein is colourle in the flask is acidic, ss. • Step 3: NaOH is added to the HCI in the flask. When the NaOH comes in contact with the solution in the flask, it turn s pink and then the pink colour quickly disapp ears. This is because the OH- from the NaOH inte ract with the phenolphthalein to change the phenolphthalein from colourless to pink. • The solution becomes cle ar again as the hydronium ions from the hydrochloric acid neutralize the added hydrox ide ions. As more NaOH is added, it takes lon ger for the pink colour to disappear. As it starts taking longer for the pink colour to disappear , the sodium hydroxide is added a dro p at a time. Acid-Base Titration 1 reached • The equivalence point of the titration isonium and when equal numbers of moles of hydr hydroxide ions have been reacted. of the • When this happens in this titration, the pH the and 7.0 is flask the in tion solu phenolphthalein indicator is colourless. possible. • Step 4: Add as little excess NaOH as to the NaOH of drop le sing a We want to add the colourless solution in the flask and have while solution in the flask turn pink and stay pink led. swir are flask the of ents cont the ator is • This permanent colour change in the indic the known as the endpoint of the titration and . over titration is To Solve the problem write the equation for the reaction: HCI(aq) + NaOH(aq) —* NaCI(aq) + H20(l) • lt • d 2 n solve for the amount of moles of the N8OH mcI titrant used, NaOH 0150 molIL x 0.02567 Lo 3.85x 10-3 mol / 3 using stoichiometry, solve for the • mole concentration of HCI knowing it is a 1:1 Foi.md wi titration experiment Tftraon 100oL I B OOI LMC C 1 2 0 0.06 0.1 015 6.2 Moles NGH u1dod SAMPLE PROBLEM 2 • In an acid-base titration, 17.45 mL of HNO were , 0.180 M nitric acid, 3 completely neutralized by 14.76 mL of Al(OH) Calculate . aluminium hydroxide, 3 aluminium the of the concentration e. hydroxid , ratio 3.85 x 10-3 mol= 0.192 M 002000 L SAMPLE ANSWER 2 : The balanced equation for the reaction ) 3HN03(aq) + Al(OH)3(aq) — A1(N03)3(aq + 3)120(i) is: The number of moles of nitric acid used x 10-3 mcI HNO3 y mci = 0.180 mci/I x 0.017451 = 3.14 number of moles of From the stoichiometry of the reaction, the ah,imwuum hydroxide reacted is: 1.05x 10-3mol )3= 3,14x 10-3molHNO3x 1 mc AliOH 3 mcI 8N03 tum hydroxide is: Therefore, the concentration of the akimin 1.05 x 10-3 mcI AtOH)3 0.0711 M 0.01476 1 2