Investigating solutions (pure or mixture) & their titration Learning objectives:
ID: 493179 • Letter: I
Question
Investigating solutions (pure or mixture) & their titration Learning objectives: Perform calculations to relote the pH and poH of a solution, the K and pr of the ocid or base and the concentrations of all species present in solution. and analyze theoretical and experimental titration curves to determine the identity and concentration of the species being titrated To understand a titration experiment, is important to understand acids & bases in Lots useful information can be extracted from an experimental one knows how it titration curve if works and what meaning of the different features. Consider the picture shown below. It starts with a solution containing 04 mol of a weak acid HA. A NaoH solution (containing oH ions) is slowly added; the oH ions wil react with HA decreasing the amount of HA present and increases the amount of A present in solution. After the first addition of NaoH: HA OH A- H29 0.4 mol 0.1 mol "Initial" (what is mixed) -0.1 mol -0.1 mol 40.1 mok 2 (going to completion) Change End" (what remains after 0.3 mol "0 mol -0.1 mol the first addition is completed) Additional information: Each icon represents 0.1 mol of a given species. The K, of this "HA" is 1.0x 104. The initial OH concentration (in burette) is 0.1 mol/15 ml or 6.7 c, pH pH Initial 50mL 80mL 110mL Vtetal 35mL. Vtetal 20mL. the pH of (a) the first, (b.) third and (c.) fourth solutions shown above. (for extra 1. do the others as well) practice, Hints: Use the pictures to know what species is in the solution. What type of solution do you have? Do you predict the pH to be neutral/acidic/basic? To find the pH, do you need to set up an ICE table? Is there another method valid for this specific solution? You may have to use the relationships between K, and pK, or between Ka and Ko..... 2. Based on your answers for Q1, sketch what the titration curve should look like.Explanation / Answer
1st beaker: moles of HA = 4*0.1 = 0.4 mol
Volume = 20 mL= 0.020L
Hence concentration of HA = moles of HA / Volume = 0.4 mol / 0.020L = 20M
-------------- HA(aq) + H2O <----> A-(aq) + H3O+ Ka = 1.0x10-4
initl.con(M): 20 ---------------------- 0 ---------- 0
eqm. con(M): (20 - y) -------------- y ---------- y
Ka = 1.0x10-4 = [A-(aq)]x[H3O+] / [HA(aq)] = y2 / (20 - y)
=> y2 + 1.0x10-4y - 2.0x10-3 = 0
=> y = 0.04467 M
=> [H3O+] = y = 0.04467 M
=> pH = - log[H3O+] = - log(0.04467) = 1.35 (answer)
4th beaker:
[A-(aq)] = moles of A-(aq) / Vt = 0.4 mol / 0.080L = 5 M
From the figure it is clear that all of the HA is converted to A-(aq). Now we can calculate the pH from salt hydrolysis formula which is
pH = (1/2)*[pKw + pKa + log[A-(aq)]]
=> pH = (1/2)x[14 - log(1.0x10-4) + log5]
=> pH = (1/2)x[14+4+0.699] = 9.35 (answer)
5th beaker:
Total moles of OH-(aq) = 2*0.1 = 0.2 mol
total volume, Vt = 110 mL = 0.110 L
[OH-(aq)] = 0.2 mol / 0.110L = 1.82 M
Hence pOH = - log[OH-(aq)] = - log(1.82) = - 0.260
=> pH = 14 - (-0.260) = 14.260 (answer)
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