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Chemistry 79 Online
OpenStudy (anonymous):

Assume that your initial aqueous solution contains only 0.100 mol of acetic acid and has a volume of 100 mL. To this acetic acid solution you will add a 1.0 mol/L NaOh solution. Let's see what happens after adding different amounts of NaOH. Suppose you have added a total of 27.0 mL of NaOH solution since the beginning of the titration. How many mol of acetate ions are present in the reaction mixture at this point?

OpenStudy (koikkara):

Alright, i will just try to help you with my small brain... 1) So your initial aq solution has 0.100 mol of acetic acid and, (1.0 mol/L) x 0.0270 L NaOH = 0.027 mol of NaOH. In this case the NaOH is the limiting reagent and the ratio of NaOH -> CH3COONa is 1 mol : 1 mol. So you would have 0.0270 mol of acetate created in this mixture. 2) First of all you started with 100ml, and then added another 100ml so the total volume is 200ml = 0.200 L. Here we have 0.100 L x (1.0 mol/L) = 0.10 mol of NaOH, which will create 0.10 mol of acetate in the reaction. So concentration of acetate is 0.10 mol / 0.200 L = 0.5 mol / L of acetate. 3) The total volume here is also 0.200 L. Since 0.10 mol of acetic acid reacted with 0.10 mol of NaOH (ratio is 1:1), and we knew that there was only 0.100 mol of acetic acid in the original solution, there is no acetic acid left. I guess that's the point of titration.. to find the point of neutralization. @ftooma

OpenStudy (koikkara):

Hope you can understand me... lol Nice to meet you !

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