experiment 27 formal lab report

experiment 27 formal lab report - Margaret McGraw April 5...

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Margaret McGraw April 5, 2016 Section 3 Experiment 27 Formal Lab Report Abstract The objective of this experiment was to determine the solubility- product constant, or the K sp , of the solid Ca(OH) 2 through the process of titration. By titrating an aqueous solution of Ca(OH) 2 , which is a strong base, with HCl, the concentration of the OH - ions in the aqueous solution of Ca(OH) s could be found because at the equivalence point the moles of H + ions present is equal to the moles of OH - ions present. Once the concentration of OH - was found, the concentration of Ca 2+ ions could be derived from the molar ratio from the equation: Ca(OH) 2 (s) + H 2 O  Ca 2+ (aq) + 2OH - (aq) Therefore, the concentration of [Ca 2+ ] was equal to half of that of the concentration of OH - . Once these values were calculated, the K sp value for each trial was calculated from the equation: K sp = [Ca 2+ ] x [OH - ] 2 The following K sp values for Ca(OH) 2 were 3.93 x 10 -5 , 3.71 x 10 -5 , 3.40 x 10 -5 . Compared to the literature value of 5.5 x 1o -6 , it can be inferred that there were systematic errors present in the experiment or titration of the Ca(OH) 2 solution. These errors ragned from the filtration of the saturated solution of Ca(OH) 2 , to the determination of the pH throughout the titration, and the analysis of the titration curve. Although the experimental values strayed from the literature value, the small value of the standard deviation, which was 4.73x 10 -12 , proved that the experimental collection of data was precise, but not accurate. Therefore, to promote accuracy in the determination of the K sp value of Ca(OH) 2 , the experiment could be modified in order to ensure that the saturated solution of Ca(OH) 2 was filtered completely and the pH value at each titration point could be collected through the Vernier rather than it being determined by oneself. Introduction An ionic compound is composed of ions connected by ionic bounding. If added to water the salt would dissociate into its ions, which in other words means that the soluble was soluble in the aqueous solution. For solids that are slightly soluble in solution, there exists an equilibrium between the solid and its aqueous ions, such as: A x B y (s) = xA y- (aq) + yB x- (aq) (Equation 1) The equilibrium constant, K c , for this reaction is: [A y- ] x [B x- ] y [A x B y ] (Equation 2) The molar concentrations of the ions, found in the numerator, ultimately depend on the amount of the solid A x B y present, which stays constant. As a
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result of this, the concentration of the solid can be combined with that of K c to create the solubility-product constant: K sp = [A y+ ] x [B x- ] y (Equation 3) The solubility product constant determines the amount of the ions that can be present in the solution. If the product of the concentrations of A y+ and B x exceed - , the solid A x B y would precipitate until the K sp I is reached once again.
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