Presentation on theme: "Buffers 1987 A. The Question NH 3 + H 2 O NH 4 + + OH- Ammonia is a weak base that dissociates in water as shown above. At 25°C, the base dissociation."— Presentation transcript:
The Question NH 3 + H 2 O NH 4 + + OH- Ammonia is a weak base that dissociates in water as shown above. At 25°C, the base dissociation constant,K b for NH 3 is 1.8 x 10^-5. Determine the hydroxide ion concentration and the percent dissociation of a.15M solution of ammonia at 25°C. Determine the pH of a solution prepared by adding.05 mole of a solid ammonium chloride to 100mL of a.15M solution of ammonia If.08 mole of solid magnesium chloride is dissolved in the solution prepared in part (b) and the resulting solution is well stired, will a precipitate of Mg(OH) 2 form? K sp for MgCl 2 is 1.5 x 10^-11.
Part A The K b expression for this problem is: K b = [NH 4 +][OH-]/[NH 3 ]. Because all the substances all have a one-to-one mole ratio, changes in the concentration of each are the same. Therefore the constant can be written as: 1.8 x 10^-5 = (x)(x)/(.15-x) When x is solved for, the value is found to be.0016M. This x value represents both the concentration of NH 4 + and OH-.
Part A continued The percentage dissociation is the amount ammonia dissociated divided by the total amount of ammonia. Because the substances are in a one-to-one mole ratio, the amount of OH- formed is equal to the amount of ammonia dissociated. Therefore the percentage can be found with: [OH-]/.15M x 100 = percentage. Plugging in the value of [OH-] we find the percentage dissociation is about 1.1 percent.
Part B In order to solve part B of the problem we must first determine the change in concentration that adding the ammonium chloride would cause. Ammonium is the only important part of the compound as it is the only part involved in the reaction. Because it will dissociate completely into one ammonium and one chloride ion, the number of moles of ammonium added is the same as the number of moles of ammonium chloride.
Part B continued Because molarity is the number of moles divided by volume in liters, we must take the number of ammonium ions added,.05 moles, and divide it by 100 mL, the amount of solution it will be added to. We find.05 moles ammonium/.1 L =.5M The change in molarity is then.5M
Part B continued The base dissociation constant is still expressed as: K b =[NH 4 +][OH-]/[NH 3 ] where the changes in concentration are equal because all substances have a one-to- one ratio.
Part B continued The concentration of NH 4 +, however, is greater by.5M so the new expression can be written as: 1.8 x 10^-5 = (x+.5)(x)/(.15-x) where x+.5 represents the concentration of ammonium and x represents the concentration of hydroxide.
Part B continued X is then solved to be equal to 5.4 x 10^-6. We can now use the solved value of the concentration of hydroxide to find the pH of the solution. We use the concentration of hydroxide to fine the pOH of the solution.
Part B continued Both pH and pOH are found by taking the negative log of the concentration of H+ ions or the concentration of OH- ions, respectively. So then pOH = -log[OH-] = -log 5.4 x 10^-6 = 5.27 Because the value of pH and the value of pOH in any solution always adds up to 14, 14-pOH = pH and thus The pH is 14-5.27 = 8.73
Part C In order to solve this problem we must know the concentrations of both OH- and Mg+. We have already solved for the concentration of OH- as 5.4 x 10^-6 Because MgCl 2 dissociates completely into one magnesium ion and two chloride ions, the amount of magnesium will be equal to the amount of magnesium chloride.
Part C continued The problem states that.08 moles of MgCl 2 is dissolved in the solution in part B which has a volume of.1 L. The concentration of Mg+ is then.08 moles Mg+/.1L =.8M To determine if any precipitate is formed, we must compare the value of [Mg+][OH-]^2 to the K sp with the given values of [Mg+] and [OH-].
Part C continued If the value is greater than the K sp then a precipitate will form. [Mg+][OH-]^2 = (.8)(5.4 x 10^-6)^2 = 2.33 x 10^-11. This value is greater than the given value of K sp for MgCl 2 and so a precipitate will form.