Experiment 6 Titration II – Acid Dissociation Constant
Figure 4 Titration curve of weak diprotic acid by NaOH(aq) Pre-Lab Notebook: Provide a title, purpose, CH3COOH / NaOH reaction, brief summary of the procedure, and table of reagents (NaOH and CH3COOH) UCCS Chem 106 Laboratory Manual Experiment 6
Acid-Base Titration Curves Using a pH Meter
1) hydrochloric acid, HCl(aq) with sodium hydroxide, NaOH(aq); 2) acetic acid, CH 3 COOH(aq) with sodium hydroxide, NaOH(aq) The recorded volume and pH values will generate titration curves that will be used to compare features of the strong acid curve versus the weak acid curve You will determine the equivalence point volume and pH for both
FINDING THE UNKNOWN MOLARITY OF ETHANOIC ACID IN VINEGAR
Through this equation, we can say that the molarity of NaOH and the molarity of CH3OON is equal since their ration is 1:1 Since the NaOH is a standard solution, it reacts with the Acetic Acid (CH3COOH) The formula to find the moles and the molarity: This means that 1M of NaOH means that there is 1 mole in the NaOH/L Tjia 3
We discussed strong acid-strong base titrations last semester
the NaOH so, how much NaOH was added? 0 013 L NaOH soln x 0 01 mol NaOH = 0 0013 mol NaOH 1 L NaOH soln one mole CH 3CO 2H consumed and one mole of CH 3CO 2– formed for every mole of OH– added CH 3CO 2H OH– CH 3CO 2– start 0 0025 0 0013 0 rxn – 0 0013 – 0 0013 + 0 0013 after rxn 0 0012 ~ 0 0 0013
E17 Acid Ionisation Constant of Acetic Acid from Titration
addition of some NaOH, the +curve is relatively flat where [H (aq)] is given by equation (2) Near the equivalence point, the pH rises sharply At the equivalence point, all of the CH 3 COOH has reacted and the solution contains CH 3 CO 2 The pH at this point is greater than 7 as the acetate ion is a weak base and undergoes hydrolysis
Acids and Bases - University of Sydney
• Common strong bases are: NaOH, KOH, Ca(OH)2, Ba(OH)2 Weak indicates an equilibrium exists between the ions and the undissociated compound in solution with, very often, the undissociated compound dominating The pH of this solution may only be calculated if the equilibrium constant, K, as well as the concentration of the starting material is
Buffer Solutions - Cal State LA We Are LA
6 Buffer Solutions ν Let’s go back to problem of adding HCl to buffer solution: ν We can use H-H eqn to make the calculations much easier [CH3COOH] = 0 100 + [HCl]added [CH3COO-] = 0 100 – [HCl]
Acid-Base Chemistry
3O+] in a 0 053 M NaOH solution Step 1: since NaOH is a strong base, dissociation is complete ∴ [OH-] = 0 053 M Step 2: Use K w to calculate [H 3O+] K w = [H 3 O +][OH-] = 1 0 x 10-14 [H 3 O +] = K w [OH-] = 1 0 x 10-14 0 053-13= 1 9 x 10 M
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139Titration
We discussed strong acid-strong base titrations last semester.Say we titrate HCl with NaOH
At the beginning of the titration, the pH is determined be the concentraion of the acid. After the end point of the titration, the pH is determined by the concentration of the base.At the end point of the tirtation what is the pH?
At the endpoint the moles of HCl = the moles of NaOH so all that is present is H2O, Cl-, and Na+. So, the pH is 7.
But what is different if a weak acid is titrated with a strong base? Lets say we are titrating a solution of acetic acid, CH3CO2H, with
sodium hydroxide, NaOH. the reaction is CH3CO2H(aq) + OH-(aq) --> CH3CO2-(aq) + H2O(l)
The sample of acetic acid is 25 mL of a acetic acid solution. It will be titrated with NaOH.What is the pH before any base has been added?
just an equilibrium problem... CH3CO2HH+CH3CO2-
init0.10~00 change- x+ x+ x equilibrium0.10 - xxx1401.8 x 10
-5 = x2 (0.10 - x) x = 0.0013 [H +] = 0.0013pH = 2.87What is the pH of the solution at the end point?
At the endpoint mol of NaOH = mol CH
3CO2H. What is present in the solution?
Na +, and CH3CO2-, (the OH- consumed the H+) Na + is neither an acid nor a base, but acetate is a base! So the pH is not 7! We have a base equilibrium problem.So, what is the concentration of the base?
How many moles of acetate are present?
One mole of acetic acid produce one mole of acetate; so,0.025 L CH
3CO2H x 0.1 mol CH 3 CO
2 H = 0.0025 mol CH
3CO2H1 L soln
0.0025 mol CH
3CO2H x 1 mol CH 3 CO
2 H = 0.0025 mol CH
3CO2- 1 mol CH
3CO2- So, what volume is the acetate in? Not 25 mL you added a solution of NaOH. how much base did you add? 1 mol acid for 1 mol base0.0025 mol CH
3CO2H x 1 mol NaOH = 0.0025 mol NaOH 1 mol CH
3CO2H141where did the NaOH come from? a NaOH solution
0.0025 mol NaOH x 1 L of solution = 0.025 L NaOH 0.1 mol NaOH
so, total volume is total vol = vol acid + vol titrant vol = 0.025 L + 0.025 L = 0.050 LSo, the concentration is
0.0025 mol CH
3 CO
2 -
= 0.050 M0.050 L
Now finally the equilibrium problem!
CH3CO2-CH3CO2HOH-
init0.0500~0 change- x+ x+ x equilibrium0.050 - xxx K b = x2(0.050 - x) remember K w = KbKa10-14 = Kb(1.8 x 10-5) K b = 5.56 x 10-105.56 x 10
-10 = x2 (0.050 - x) small x approximation x2 = 2.78 x 10-11
x = 5.27 x 10 -6 [OH -] = 5.27 x 10-6 so, [H [H +][OH-] = 10-14 142[H+] = 10-14/(5.27 x 10-6) [H +] = 1.90 x 10-9 pH = 8.72 Say we stop the titration at 13 mL NaOH. What is the pH? What is in solution after 13 mL NaOH have been added?