Experiment 5, pH titration of acetic acid with sodium carbonate, Calculation of the concentration of the unknown acid. I've got a summary of the results I've gotten from this in previous labs right down here. I know from Experiment 4 that the solution I made is 0.893 mole or sodium carbonate. From my graphical determination of the equivalence point, I have determined that I needed 15.10 milliliters of sodium carbonate to get there. And my acetic acid solution, my vinegar that I used, I used 10 amounts of 2 milliliters, which for me each one was 2.01, so that amounts to 20.10 milliliters. Now, at the equivalence point, my moles of acid equals my moles of base. And both of these have 1's next to them that are implied, and that's because both of them have 1's next to them in the balanced chemical reaction for this process. Now, from there, I will start because I have my volume and my malarity of my base. I can find my moles of base. I'm going to convert my milliliters into leaders. And multiply that by the malarity, which is really 0.893. Moles of sodium carbonate per liter. When I do that, I get 0.01348, 4.3 moles sodium carbonate. And I only have 3 sig figs because I'm limited by my malarity. This moles of sodium carbonate is my moles of base. It is equal to my moles of acid, so now I will write equals this number, moles acetic acid. And that's the beauty of the equivalence point, as I can say, these 2 numbers are equal to each other. Now, because I have the moles of acetic acid, and I have the volume of acetic acid, I can now find the malarity of the acetic acid by taking moles divided by volume and leaders. So my concentration of acetic acid is going to be moles divided by leaders. And if I multiply this out, I have my number of moles already in my calculator divided by 0.02.010. I get 0.67086 malarity acetic acid as my result.