• Author / Uploaded
• Kendy Livi Danawati

Citation preview

A 50.00 mL aliquot of a solution containing iron(II) and iron(III) required 13.73 mL of 0.01200 M EDTA when titrated at pH 2.0 and 29.62 mL when titrated at pH 6.0. Express the concentration of the solution in terms of the parts per million of each solute. Trivalent cations form more stable complexes at pH around 1 than divalent cations At very low pH, divalent cations form unstable complexes Additional information: Fe3+ is more acidic than Fe2+ pKa of Fe3+ = 2.18 and pKa of Fe2+ = 9.39 At pH 2.0, only Fe3+ stable complexes may form At pH 6.0, both Fe2+ and Fe3+ complexes may form At pH = 2.0 Moles Fe3+ = moles EDTA = (0.01373 L)(0.01200 M) = 1.648 x 10-4 mol Concentration Fe3+ = 1.648 x 10-4 mol/0.05000 L = 3.295 x 10-3 mol/L = 184.0 mg/L At pH = 6.0 Moles Fe3+ + Fe2+ = moles EDTA = (0.02962 L)(0.01200 M) = 3.554 x 10-4 mol Moles Fe2+ = 3.554 x 10-4 mol - 1.648 x 10-4 mol = 1.906 x 10-4 mol Concentration Fe2+ = 1.906 x 10-4 mol/0.05000 L = 3.812 x 10-3 mol/L = 212.9 mg/L A 50.00 mL sample of a white dinner wine required 21.48 mL of 0.03776 M NaoH to achieve a phenolphthalein end point. Express the acidity of the wine in terms of grams of tartaric acid (H2C4H4O6; 150.09 g/mol) per 100 mL. (Assume that two hydrogens of the acid are titrated.) Find moles of OH it takes to reach the endpoint (color change, equivalence point): 21.48 mL * (0.03776 mol OH-/1000mL) = 0.000811085 mol OHSince it takes twice the moles of NaOH to neutralize each mole of tartaric acid (since there are 2 moles of H+ in each tartaric acid molecule), 0.0008110085 moles of Base will neutralize half or 0.000405504 moles of Acid. Convert this equivalence of acid you have found into grams using the MW: 0.000405504 moles of H+ * (150.09 g/mol) = 0.06087 grams of H+ Now convert grams to concentration of grams/100 mL using the total volume of all the wine that you started with: 0.06087 grams/50 mL = 0.1217 grams tartaric acid/100mL Basically you are using the act of adding a Base painstakingly measuring every single drop until

you reach endpoint (neutral pH), then using that amount to go backward and calculate how much acid was in there to start with.

Source: College chemistry (Gen chem, organic chem, biochem etc) Titration of a 0.7439g sample of impure Na2B4O7 required 31.64 mL of 0.1081M HCl. Express the results of this analysis in terms of percent Na2B4)7*10H2O. HELP! Na2B4O7 + 5H2O + 2HCl --> 4 H3BO3 + 2NaCl 0.03164 L @ 0.1081mol / litre HCl = 0.003420 moles HCl 0.003420 moles HCl @ 1 mol Na2B4O7 / 2 mol HCl = 0.001710 moles borax =============== 0.001710 mol Na2B4O7*10H2O @ 381.37 g/mol = 0.6522 grams of borax % = 0.6522 g borax / 0.7439g sample = 87.67 % your answer = 87.67 % Na2B4O7*10H2O Treatment of Hydroxylamine (H2NOH) with an excess of Fe (III) results in the formation of N2O and an equivalent amount of Fe (II): 2 H2NOH + 4Fe (III) ---> N2O (g) + 4Fe (II) + 4H + H2O Calculate the molar concentration of an H2NOH solution if the Fe (II) produced by treatment of a 50.00 ml aliquot required 23.61 ml of 0.02170M K2Cr2O7. moles Cr2O72- = 0.02361 L x 0.02170 M= 0.0005123 6 Fe2+ + Cr2O72- + 14 H+ = 6 Fe3+ + 2 Cr3+ + 7 H2O moles Fe2+ = 6 x 0.0005123 =0.003074 moles H2NOH = 0.003074 /2= 0.001537 M = 0.001537 / 0.0500 L= 0.03074