Tutorial on Equilibrium Calculations: Here are a few examples of a equilibrium problems that may help explain them. Consider acetic acid HAc with a pKA = -4.75 (or Ka=1.7610-5). HAc H + Ac [H ][Ac ] 1.76 105 [HAc] 1) If we have a solution of 0.1M HAc what is the pH? Before equilibrium has set in we have only 0.1 M HAc then we let equilibrium take place and x moles of the HAc dissociates. HAc H+ Acstart 0.1 M 0 0 At eq 0.1 – x x x We start with the 0.1M of acetic acid but at equilibrium it dissociates and so you lose x of it and gain x for both [H+] and [Ac-]. Writing the equilibrium constant you get [H ][Ac ] [x][x] K 1.76 105 [HAc] [0.1 x] To solve this you can ignore the x in the denominator and you get x=[H+]=1.32x10-3 M or a pH =-log[H+] = 2.88. K Now try this 2) You have a solution that you add 0.1 M HAc and 0.2 M NaAc. What is the pH? Preceding as above we get HAc H+ Acstart 0.1 M 0 0.2 M At eq 0.1 – x x 0.2+x Note that the reaction must proceed to the right since we say there is no H+ (actually there is a little from the dissociation of water but that is negligible in this problem).. We get K [x][0.2 x] 1.76 105 [0.1 x] [x][0.2] 1.76 105 [0.1] + -6 solving we get x=[H ]=8.8 x 10 ; pH=5.06. The pH increased since we added a weak base NaAc Now x is small compared to both 0.1 and 0.2 so we get K 3) Now try a strong acid and a weak base. You have a solution of HCl and the salt of a weak acid NaAc. [HCl] =.1 M; and NaAc = .12 M HAc H+ Acstart 0.0 M 0.1 0.12 M At eq x 0.1-x 0.12-x [0.1 x][0.12 x] 1.76 105 [x] Unfortunately x is not small compared to 0.1 or 0.12 so we can not ignore it in this equation. It can be solved but is takes a quadratic. Multiplying out and rearranging and we get 0 1.2 102 (0.22 1.76 105 )x x 2 if we ignore the 1.76x10-5 compared with 0.22 we get x = 0.1 M! This is strange since it implies that [H+] = 0.1 – x = 0.0 Now we get: K The problem with this answer is due to round off error. If you do it exactly (i.e. don't ignore 1.76x10-5) you get two roots for x of 0.1201 and 0.0.09991. The first answer (0.1201 is physically impossible since you do not start with enough Ac to give that much HAC) Then [H+] = 0.1 –x =0.1 – 0.09991 = 0.00009. However these numbers are really outside of the error in our starting concentrations since you need 5 digits to do the last subtraction. So we need to do something else. Assume that the reaction precedes to the left as far as it can go and you form 0.1 M HAc and are left with no H+ and only 0.02 M Ac-. An alternative way to look at it is rather then starting by mixing 0.1 HCL and 0.12 NaAc you add 0.1 M HAc and 0.02 M NaAc and 0.1 M NaCl. You get the same chemical solution no matter which things you added originally since they all added up to the same amounts of H, Cl, Na and Ac and equilibrium allows the various species to form as the equilibrium constant requires. This second way we have HAc H+ Acoriginal start 0M 0.10 0.12 M rex moved to left 0.01M 0.0 0.02 M At eq 0.1 - x x 0.02 + x Now we get [x][0.02 x] K 1.76 105 in this case x is small compared to 0.02 so we solve and [0.1 x] get [x][0.02] K 1.76 105 and x = [H+] =8.8e-5 or pH = 4.06. [0.1] So the lesson is sometimes in equilibrium problems it helps to push the equilibrium all the way one way and them let it come back rather then just using the original starting species. A more complicated problem: Telluric acid is a weak acid, H2TeO4. Here is a problem to help with the HW. Assume you have a beaker of 0.1M HCl. You are going to add pinches of Na2TeO4 to the beaker and each pinch you want to know the pH. Each pinch will contain 0.03 M of Na2TeO2 ( I will just call it Na2T). In the lab of course you would grab a pH meter and measure the pH but we will calculate it. The equilibrium is H 2T H HT pK1 6.27; K1 5.4 107 HT H T 2 pK 2 8.43; K 2 3.7 109 What we will do is calculate the pH at the start. At start [H+] = 0.1M (HCl is strong acid and completely dissociates). Thus pH = -log([H+])=1 Now add the pinch of Na2T and the salt dissociates so you have Na2T 2Na T 2 start [HT-] 0 [H+] 0.1 [T2-] 0.03 Lets let the reaction go all the way to the left and then come back for equilibrium: [HT-] [H+] [T2-] start 0 0.1 0.03 shift left 1 0.03 0.07 0 However we still have more H+ that will then protonate the HT[HT2] [H+] [HT-] shift left 1 0 0.07 0.03 shift left 2 0.03 0.04 0 eq 0.03-y 0.04+y y Now we get: K1 5.4 107 4.1107 y [H ][HA ] (0.04 y)(y) .04 y [HA2 ] 0.03 y .03 where we guessed that y was much smaller than either 0.04 or 0.03 and so could ignore it when compared to both of them. This was then an easy equation to solve. and the pH is pH log([ H ]) log(. 04) 1.4 Note that here since in the second equilibrium we don't use up all the acid it pH can be calculated without actuualy solving the above equation since H+ is just 0.4 the left over acid. Now add 0.03 more Na2T. start shift left 1 [HT-] 0 0.06 [H+] 0.1 0.04 However we still have some acid that will then protonate the HT[HT2] [H+] shift left 1 0.0 0.04 shift left 2 0.04 0.0 eq 0.04-y y [T2-] 0.06 0 [HT-] 0.06 0.02 0.02+y Now y is the H+ concentration and we get [H ][HA ] (y)(0.02 y) .02y K1 5.4 107 [HA2 ] 0.04 y .04 1.08 106 y and pH log([ H ]) log(1.08 106 ) 6.0 Do this again and what do we get [HT-] start 0 shift left 0.09 [H+] 0.1 0.01 However we still have some acid that will then protonate the HT[HT2] [H+] shift left 1 0.0 0.01 shift left 2 0.01 0.0 eq 0.01-y y Now y is the H+ concentration and we get [H ][HA ] (y)(0.08 y) .08y 7 K1 5.4 10 [HA2 ] 0.01 y .01 6.8 108 y and pH log([ H ]) log( 6.8 108 ) 7.2 [T2-] 0.09 0 [HT-] 0.09 0.08 0.08+y A general approach to equilibrium problems. The most general approach to equilibrium problems is to write n equations in n unknowns and solve them. This is not as hard as it sounds since the equations are easy to write down. They include three classes of equations, conservation of matter, charge neutrality, and equilibrium constants. An example is the above problem of a solution made up of NaAc 0.12M and HCl 0.1M. We know that NaAc and HCl completely dissociate. We have by the conservation of Ac- that Total Ac = [Ac] + [HAc] where the Total Ac is just 0.12 M Charge neutrality requires that the number of negative and positive ions be equal: + [H ] + [Na+] = [Ac-] + [OH-] +[Cl-] where [Na+] = 0.12 M and [Cl-] = 0.1 M. The equilibrium constant equations are: [H ][Ac ] KA [HAc] K w [H ][OH ] or listing all together and plugging in the concentrations that we know: 0.12 [Ac ] [HAc] [H ] 0.12 [Ac ] [OH ] 0.1 KA [H ][Ac ] [HAc] K w [H ][OH ] We have 4 equations and 4 unknowns. To solve we approximate the [OH-] concentration as much less then 0.1 M, so we drop that term in the second equation and solve for [Ac-]. 0.12 [Ac ] [HAc] [Ac ] [H ] 0.12 0.1 [H ] .02 KA [H ][Ac ] [HAc] K w [H ][OH ] using this value in the first and the third equations we get 0.12 [H ] .02 [HAc] KA [H ][H ] .02 [HAc] now plugging the first into the second for [HAc] we get: KA [H ][H ] 0.02 0.12 [H ] This equation is the familiar one from above that we solved by assuming that [H+] is small compared to 0.02 or 0.12.