Consider the titration of 44.2mL of 0.240M HF with 0.180 M NaOH . Calculate the
ID: 606064 • Letter: C
Question
Consider the titration of 44.2mL of 0.240M HF with 0.180 M NaOH . Calculate the pH at each of the following points. A)How many milliliters of base are required to reach the equivalence point? B)Calculate pH after addition of 11.1mL of base C)Calculate the pH at halfway to equivalence point D) Calculate pH at equivalence pointExplanation / Answer
Consider the titration of 35.2 mL of 0.235 M with 0.215 M NaOH. Calculate the pH at each of the following points: a)after the addition of 8.80 mL base b) halfway to the equivalence point c) at the equivalence point d) after the addition of 70.4 mL of base The answer above is correct if you're titrating a strong acid, but check out the procedure below if it's a weak one. It's good to practice doing the actual calculations for these. I can't find the numbers anyway since I don't know what acid you have .235 M of. I'm assuming it's a weak acid. Here are the steps you should follow: First, calculate the initial pH. Use an ICE table to find the pH of the initial acid solution (the .235 M of weak acid sitting in water, without any base added). Use the reaction HA + H2O --> A- + H3O+. Initially you have a .235 M solution of acid. Look up the Ka of the acid and set up the equilibrium constant Ka = [A-][H3O+]/[HA] - x. This is equal to Ka = x^2/[HA] because you can assume x is small if the equilibrium constant is much less than one (look this rule up in your textbook and be careful; you'll have to use the quadratic equation if x is anywhere close to Ka), and the concentration of HA lost is converted to A- and H3O+ in equal amounts. Once you find x, you have your hydronium ion concentration, so take the -log of this to get initial pH. I know you didn't require this here, but it's important to be able to compare it to the other points. for a), 8.80 mL NaOH * (.215 mol OH-/1000 mL NaOH) = 1.89 * 10^-3 mol OH- added. Figure out how many moles of acid you had initially: 35.2 mL HA * (.235 mol HA/1000 mL HA) = 8.27 * 10^-3 mol HA. So when you add this much OH-, you neutralize an equal amount of HA (converted to water and A-) and are left with about 6.38 * 10^-3 mol HA and 1.89 * 10^-3 mol A-. Use the Henderson-Hasselbalch equation (pH = pKa + log [A-]/[HA]) to get the pH at this point. Remember that you need to divide the moles of HA and A- by the new volume (.0352 L + .008 L) to get their concentrations before using HH. b) At the half-equivalence point, half the original quantity of HA has been neutralized by added OH-, so you have equal concentrations of HA and A- (equal numbers of moles in an equal volume). According to Henderson-Hasselbalch, pH = pKa + log (1/1) = pKa + log 1 = pKa + 0 = pKa. So pH at half-equivalence is always equal to the pKa of the weak acid (-log of the Ka). c) At equivalence, you have neutralized all the HA by adding an equal number of moles of OH-. Thus all HA has been converted to A-, so you have 8.27 * 10^-3 mol A-. You had to add 8.27 * 10^-3 mol OH- * (1 L/.215 mol OH-) * (1000 mL/1 L) = y mL of OH- to get to equivalence. Add this to the original volume of 35.2 mL then divide the 8.27 * 10^-3 mol A- by this total volume. Make another ice table with only A- present initially: A- + H2O --> HA + OH-. Use the base dissociation constant Kb and do a calculation analogous to the Ka ice table. Your x will now be [OH-]. -log [OH-] = pOH. Now pH + pOH = 14 always, so pH = 14 - pOH. You can ignore the dissociation of HA here because it is a weak acid, and the H3O+ it produces has little effect on the pH compared to the relatively large amount of OH- produced. d) After the equivalence point, pH depends on the amount of OH- present. The contribution of A- to the amount of OH- can be ignored because it is a weak base and the amount of OH- it produces is small compared to the amount you are adding (there is no acid left to neutralize the OH- added after equivalence). However, you do need to add the moles you are adding to the amount present at equivalence (calculated in part c above; convert x, the concentration, into moles of OH-). When you have added 70.4 mL of base, this means you have added (70.4 mL - volume at equivalence) past the equivalence point (70.4 means compared to the zero initially, not 70.4 in addition to equivalence). Convert the volume added past equivalence into moles of OH-, then divide these moles by the new concentration (volume at equivalence + volume NaOH added past equivalence) to get the concentration of OH- in solution. Then take -log [OH-] = pOH, and 14 - pOH = pH. EDIT: I can't remember whether adding the moles present at equivalence to those added after will make a big enough difference to appear in your significant figures, but maybe try it with and without adding the moles from A- and see which one gives you the right answer. Sorry if this looks long and tedious, but really understanding this procedure and the reasons behind it is important, especially if you're in college level chem. Acids and bases are one of the trickier subjects, and not completely knowing what I was doing with these kinds of problems messed me up in first semester general chem. Let me know if anything needs clarification or if this procedure doesn't give the same answers as your textbook.