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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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