tuA A Salubillty Product Constant Introduction 1he a slightly soluble ial name.
ID: 525442 • Letter: T
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tuA A Salubillty Product Constant Introduction 1he a slightly soluble ial name. It called the solubility product constant Purpose pound with its dissolved ions leads to another type constant for this kind of equilibrium has experiment, you will determine th amount constant. will also demonstrate and calculate it of that is used to that of Caooo is independent water saturated solution of this substance. the concept of the Experiment ubstance co can be calculated adopt for Ca(IO,2 in experiment. Gammon His measured, the solubility product constant for ted solution is The equilibrium 17.3). This is the method that you will between solid Callos and its ions in a If some analytical techniq is used to ue can be calculated the solubility ofCaOO) will be kno of either Ca ions or IO,ions in the own, and the solubility product constant (K In this experiment, you will determine the concentration of ions through a with a solution of lo ti as KI (potassi ium using starch an indicator. A definition of iodide), Acid-Base Titration standardization can be found in the ex product containing iodine. The reaction is react with the ions to give Las the sole The molecular iodine reacts with so ions during the titration according to where so, is the tetrathionate ion. When these two equations are combined, multiplying the second reaction by 3 so the intermediate Lis e nated, the net reaction occurring in the titration is obtained. Starch is used as an indicator in this titration because it reacts with I2reversibly to form a darkblue color. 12 is consumed in the titration, so the color fades as the titration progresses. You will know when a stoichiometric volume of the NazS20, solution has been added because at that point, one d of that solution will cause the disappearance of the last trace of the blue color. A trial titration will enable you to find the approximate volume that is required before you do the first of two exact titrations 3Congage Leaming All Rights Reserved. May not be scanned, copied or duplicated or posted a publicly accessible website in whole or inExplanation / Answer
In this iodometric titration (standardisation part), the iodate ions react with iodide ions and acidic protons to liberate iodine. This liberated iodine is then titrated with sodium thiosulfate. The balanced reactions are given as:
IO3(aq) + 5I(aq) + 6H+(aq) -----> 3I2(aq) + 3H2O(l)
I2(aq) + 2S2O3-2(aq) ------> 2I(aq) + S4O6-2(aq)
Number of molesof IO3 = concentration of IO3 solution X volume of IO3 solution
= 0.01 mol L-1 X 10 X 10-3 L
= 1.0 X 10-4 mol
From the above equation we know that 1 mol of I2 is consumed per 2 mol of S2O3-2
And 3 mol of I2 are formed per mol of IO3
Therefore, 6 mol of S2O3-2 are consumed per mol of IO3
Number of moles of S2O3-2 = 1/6 X Number of moles of IO3
= 1.0 X 10-4 / 6 mol
= 1.67 X 10-5 mol
concentration of S2O3-2 solution = Number of moles of S2O3-2 /volume of S2O3-2 solution
For Trial 1
concentration of S2O3-2 solution = 1.67 X 10-5 mol / 21.1 X 10-3 L
= 7.91 X 10-4 mol L-1
= 7.91 X 10-4 M
For Trial 2
concentration of S2O3-2 solution = 1.67 X 10-5 mol / 20.9 X 10-3 L
= 7.99 X 10-4 mol L-1
= 7.99 X 10-4 M
Mean molarity = ½ (7.91 X 10-4 M + 7.99 X 10-4 M) = 7.95 X 10-4 M
NB: What you have entered as moles of IO3 and S2O3-2 are actually the molar masses of these compounds.
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