Measurement of Ksp Values and Determining Concentration of an Unknown NaCl Solution Lab Answers

Measurement of Ksp values and Determining Concentration of an Unknown NaCl Solution Introduction Last semester in General Chemistry I, you learned solubility rules to identify combinations of ions that produce insoluble compounds. In that context, solubility was binary – either a substance was soluble or not. Those insoluble compounds are more accurately described as poorly soluble, requiring very low (but measurable) concentrations to precipitate from solution. The solubility of any ionic compound can be represented as an equilibrium between the pure solid and the dissociated aqueous ions. When the solid is placed on the reactants side of the equilibrium, the equilibrium constant is known as the solubility product, Ksp. Some examples are shown below: Ba(NO3)2 (s) BaCO3 (s) ⇌ ⇌ Ba2+ (aq) Ba2+ (aq) + + 2 NO3– (aq) Ksp1 = CO32– (aq) Ksp1 = The magnitude of Ksp gives some idea of how soluble a compound is – poorly soluble compounds have very small Ksp values. Be careful, though – here be dragons. Because Ba(NO3)2 and BaCO3 produce different numbers of ions and have different exponents in the equilibrium expression, their K sp values are not directly comparable. We can use an ICE table and Ksp for each compound to calculate its molar solubility, the molarity of a saturated solution. Initial Change Equilibrium Ba(NO3)2 (s) - ⇌ Ba2+ (aq) 0 +x x + 2 NO3– (aq) 0 +2x 2x Ksp = [Ba2+][NO3-]2 x = 0.105 M (molar solubility) 27.4 g/L For a generic ionic compound (MaXb) dissolved in water, Ksp = [ax] [bx]b. For example, Al2O3 would have the expression Ksp = (2x)2(3x)3. The 1:1 ratio of ions in BaCO3 gives the expression Ksp = x2 a molar solubility of x = M (0.00985 g/L). To put that in perspective, it would take a little over 100 L of water to a 1 CRC Handbook of Chemistry and Physics. 87th ed. CRC Press/Taylor and Francis Group, 2006; p 8118. dissolve 1 g of BaCO3. The cutoff between slightly soluble and insoluble used to establish solubility rules varies somewhat between textbooks, but this would be considered insoluble by any of them. Precipitation When you first encountered the solubility rules, you used them to predict when a precipitation reaction would occur and wrote net ionic equations for these reactions. Back then you didn’t worry about concentration; if ions that form an insoluble compound were present, precipitation occurred. Now that we know about solubility equilibria and saturated/unsaturated/supersaturated solutions, we can dive a little deeper. To determine whether a precipitate will form we must calculate the value of Q, the reaction quotient for the solubility equilibrium. If Q = K the solution is saturated, if Q > K it’s supersaturated, and if Q < K the solution is unsaturated. As we learned in the solutions chapter, a precipitate will only form for a supersaturated solution and will stop forming once the saturation point is reached (Q = K). Example: Consider mixing 25 mL each of aqueous 0.010 M Ba(NO3)2 and 0.010 M Na2CO3. Knowing that BaCO3 is fairly insoluble you might write the following molecular and net ionic equations: Ba(NO3)2 (aq) + Na2CO3 (aq) 2+ BaCO3 (s) + 2 NaNO3 (aq) 2- Ba (aq) + CO3 (aq) BaCO3 (s) 2+ Mixing the solutions dilutes each solution, so [Ba ] and [CO32-] each decrease to 0.0050 M. Using these values, we can calculate Q. Note the net ionic equation above is the reverse of the K sp equilibrium. If we’re comparing Q to Ksp , we have to use the expression for Ksp, where the dissociated ions are on the products side. BaCO3 (s) ⇌ Ba2+ (aq) Q = [Ba2+][CO32-]; + CO32– (aq) Q = (0.0050)2 Q = ; Ksp = Q > Ksp Because Q > Ksp, the solution is supersaturated and precipitate should form until Q = K. If we wanted to set know how much solid forms, we could set up an ICE table: Initial Change Equilibrium BaCO3 (s) - ⇌ Ba2+ (aq) + 0.0050 -x 0.0050 – x = (0.005 – x)2 Using the quadratic formula, x = 0.00495 M CO32– (aq) 0.0050 -x 0.0050 – x [Ba2+] = [CO32-] = 0.0050M – 0.00495 [Ba2+] = [CO32-] = M Initially, [Ba2+] and [CO32-] had the values, so the equilibrium concentrations are also equal and match the molar solubility of BaCO32-. That’s certainly not always the case. Given the 50 mL of solution and change of 0.00495M, 0.049 g of solid BaCO 3 should form. 0.050 L soln 0.0495 mol Ba2+ 1 mol BaCO3 197.34 g BaCO3 L soln 1 mol Ba2+ mol BaCO3 = 0.049 g BaCO3 Common-ion effect Like other equilibria, the solubility of an ionic compound is subject to LeChâtelier’s principle. For instance, if Na2CO3 is added to a saturated solution of BaCO3, the equilibrium will shift in response to the change in [CO32-]. Since Na2CO3 and BaCO3 each form carbonate ions when they dissociate, carbonate is a common ion and this type of shift in equilibrium as a common ion effect. In the context of solubility equilibria, the presence of common ions decreases the molar solubility of the ionic compound. While Ksp is constant at the given temperature, the molar solubility is not. Earlier, we determined that the molar solubility of BaCO3 was M but this assumed the solvent was pure water. If instead we dissolve BaCO3 in a solution containing 0.10 M Na2CO3 carbonate, our ICE table looks a bit different: Initial Change Equilibrium BaCO3 (s) - Ba2+ (aq) 0 +x x ⇌ + CO32– (aq) 0.10 +x 0.10 + x = x (0.10+x) = 0.10x + x2 x= M The quadratic formula can be used to solve for x in the calculation above, but when Ksp is small it we can make the simplifying assumption that x is negligible compared to the soluble source of common ion. In this case we can assume x << 0.10 and therefore 0.10 + x is approximately 0.1. Making that assumption here leads to the same result. = x (0.10); x= M The small x assumption is particularly useful when working with compounds that are not a 1:1 ratio of ions.. Consider the same problem but using Ag2CO3 instead of BaCO3. Ksp = (2x)2(0.10+x) The quadratic formula is no use here as the resulting is cubic (x 3) equation is a challenge to solve algebraically. The small x assumption simplifies this to Ksp = (2x)2(0.1), which is trivial by comparison. Data: 1. Determination of Ksp Values Compound formula Mass of solid added, g SrSO4 Water volume, mL [Cation], M [Anion], M Mass of solid remaining, g .5g 50 mL .0005831 .0005831 .494645 .5g 50mL .0002565 .00012824 .498232 .5g 50mL 4.5037x10^4 2 1.00488x10^ -7 .50080 2. Common Ion Effect: Determination of Unknown NaCl Volume of unknown solution, mL [Ag+], M 99.927mL 1.598437 X10^-10 Mass of AgCl(s), g 1 Data Analysis 1. Determine the molar solubility and value of Ksp for each compound in water at 25°C. Include the chemical equation for each equilibrium, the expression for Ksp (in terms of molarities and x), and show all work in calculations Compound 1 Chemical equation: SrSO4 + 2H2O-----> <------ H2SO4 +SrOH2 Molar solubility (x) = .0005831 Ksp = 3.4x10^-7 Compound 2 Chemical equation: Ag(O3)+ 2H2O -----> 2AgOH + H2CO3 <------ Molar solubility (x) = .0001282 I 0 0. C 2X. X. .0002565/2= .0001282 E 2X. X Ksp = .0001282 8.43X10^-12 = .0002565^2. Compound 3 Chemical equation: Al(OH)3 +H2O ----->Al3(s) + 3OH-(aq) <------ Molar solubility (x) = 3.3X10^-8 1.00488X10^-7/3 I 0. 0 C +X. +3X. E Y. 3X Ksp = 4.570X10^-33 . (1.00488X10^-7)^3X4.5037X10^-12 2. Determine the molarity of the unknown NaCl solution. Show all work. AgCl ----- Ag +Cl 1.83x10^-10/(1.598437x10^-10). 1.598437x10^-10) 1.15292 3. Determine the mass of NaCl dissolved in the unknown NaCl solution. Show your work. 1L/1000mL X 1.153/L. X 99.938mL. X 58.44g/mole 6.733