The solution which contain both a weak acid (here acetic acid) and its conjugate base (here CH3COO-) , then it is called a Buffer. And pH of a buffer is calculated by Henderson Hasselbalch equation
pH = pKa + log [salt]/[acid]
Now given pH of the solution = 2.40
This the [H3O+] = 10-pH = 10-2.40 = 0.00398 M
The dissociation of CH3COOH is water is
CH3COOH + H2O ------------> CH3COO- + H3O+
Now the given [H3O+] is actually at equilibrium. Thus if we consider the initial concentration of CH3COOH taken = xM
Then the ICE table will be
Reaction | CH3COOH | CH3COO- | H3O+ |
Initial | xM | 0 | 0 |
Change | -0.00398 M | +0.00398 M | +0.00398 M |
Equilibrium | xM-0.00398 M | 0.00398 M | 0.00398 M |
Now Ka for the reaction will be-
Ka = [CH3COO-][H3O+] / [CH3COOH]
Now if the given [CH3COOH] at equilibrium = xM-0.00398 M = 0.006 M, then putting the values-
Ka = [CH3COO-][H3O+] / [CH3COOH]
Ka = [0.00398][0.00398] / [0.006]
= 0.00264
Now the actual Ka value for Acetic acid = 1.8x10-5 = 0.000018
Thus the % error = (theoritical value - calculated value) / theoritical value * 100
= (0.000018 - 0.00264) / 0.000018 * 100
= [-0.002622 ] / 0.000018 * 100
= [0.002622 ] / 0.000018 * 100
= 14566 %
This indicates that our calculated Ka value in completely wrong. Since the error should be within +/- 5%
Possible source of error- wrong calculation of initial concentration
- wrong calculation of pH
- wrong calculation of CH3COOH left at equilibrium
Experiment VII: Buffers Lab Report ( 49 pts) I. Determination of the K, of acetic acid...
Need help please L. Determination of K, of acetic aci A. Measure out 10.0 mL of 1.0 M acetic acid (CH,COOH) into a beaker B. Rinse the pH probe with DI H,O and place it into the solution in the beaker eneurine thst the vin Experiment VIl: Buffers Lab Report Il. Determination of the K, of acetic acid 1. Measured pH of the solution pH 2. Calculate the [H,0'] at equilibrium for this solution. (include units) 3. Calculate the (CH,COO]...
Buffers 1. Indicate which of the following pairs of compounds could be used to make a buffer solution by placing an X in the appropriate column. Buffer Solution | Not a Buffer Solution Pair of Compounds HF and Cl H.COs and OH HNO2 and NO NHa and NH L 2. Write out the reaction of acetic acid CH.COOH reacting with water. 3. What is the pH of a buffer solution in which the (CH3COOH) is 0.229 M and the (CH...
I. REACTION EQUILIBRIUM IN NON-IDEAL SYSTEMS The acetic acid (CH:COOH) aqueous solution with a concentration of 0.1 mol/kg has a hydrogen ion concentration of 1.35x10-3 mol/kg. i. Write the acid dissociation equation. ii. Calculate the acid dissociation constant, Ka, considering the ionic activity. jïi. Calculate the pH of the acidic solution. iv. What is the new pH if the solution contains 0.1 mol CH3COOH and 0.4 mol of sodium acetate, CH3COONa, in 1.0 L of water? Consider the negligible quantitities....
1) Acetic Acid is mixed with water in an experiment. The resulting concentration of H3O+ ions in the reaction at equilibrium (after everything has been mixed) is 0.013M. What is the pH of the new solution at equilibrium? 2) A neutralization reaction with Acetic Acid and Sodium Hydroxide is run: CH3COOH + NaOH -> CH3COONa + H2O When the conjugate base (CH3COONa) concentration is 0.008M and the conjugate acid (CH3COOH) concentration in 0.0032M, what is the pH of the solution?...
Buffers Lab (I ONLY NEED HELP WITH #5) The 2 buffers are (acetic acid + sodium acetate) and (acetic acid + sodium hydroxide) 3:The two buffers that you will make in class are identical. If 5.00mL of a 1.0 M NaOH solution is added to one of these buffers, what would be the resulting pH after addition? ---Had to make a IACE table and what I got was pH=5.20 4: What is the total volume of NaOH (1.0 M) you...
I'm having trouble with the calculations for my chemistry lab: Determination of an Acid Dissociation Constant, Ka Half-Neutralization Molarity of acetic acid: 2 M Molarity of NaOH: 1 M Volume of acetic acid: 25 mL NaOH Solution Final Buret Reading: 6.23 mL Initial Buret Reading: 0.01 mL Volume added: 6.22 mL Total Volume of Solution: 250 mL pH: 3.85 Caculating Ka Initial number of moles HAn: ? OH-: ? Number of moles at equilibrium: HAn: ? An-: ? Equilibrium concentrations,...
starting with a solution at equilibrium containing .5 M acetate and .5 M acetic acid, then adding .1 M sodiun acetate, what would be the pH of the original solution and the solution after adding sodium acetate. the dissociation constant (Ka) for acetic acid is 1*10^-6
For the following dissociation of acetic acid, Ka = 1.760 x 10-5: CH3COOH + H2O ? CH3COO- + H3O+ Sodium acetate completely dissociates in solution according to the following reaction: NaCH3COO ? Na+ + CH3COO- A solution was prepared which contains CH3COOH at a pre-equilibrium concentration of 0.05782 M and NaCH3COO at a pre-equilibrium concentration of 0.04991 M. I know all answers, just do not know how to find them. What is the equilibrium concentration of H3O+? (Answer: 0.00002037) What...
Acetic acid CH3COOH 1.8 × 10-5 Acetylsalicylic acid (aspirin) HC9H7O4 3.0 × 10-4 Equal volumes of 0.506 M aqueous solutions of acetic acid (CH3COOH(aq)) and sodium acetylsalicylate (NaC,H-04) are mixed. (1) Write the net ionic equation for the overall reaction that takes place as the system comes to equilibrium. Write acetic acid, benzoic acid or formic acid and their conjugates in the form RCOOH/RCOO. For example, benzoic acid should be written "CH3COOH" NOT "CzH;CO,H". It is not necessary to include...
Acetic Acid (CH3COOH) has a Ka value of 8 x 1-5. If I have a 0.15 M acetic acid solution, what are the following values? pH pOH [H3O+] [OH-] If I add 0..15 M sodium acetate to the solution what is the pH value here? Remember that pH = pKa + log [A-]/[HA].