Equilibrium constant of the chemical reaction can be termed as the reaction quotient of that same reaction in which the froward reaction and backward reaction has attained a dynamic equilibrium.
Mathematically , it can be defined as the ratio of the concentration of the product species to the reactant species.
Chemical reaction of acetic acid dissolution can be written as -
CH3COOH (l) + H2O(aq) -------> CH3COO-(aq) + H3O+(aq)
Ka = [CH3COO-] [H3O+] / [CH3COOH] [H2O]
where Ka = dissociation constant of acetic acid.
Activity coefficient is represented by and activity multiplied by this activity coefficient makes the concentration of any species.
Given that
pKa = 10-4.7
As we know that pka = -log ka , Thus 10-4.7 = - log Ka, taking anti logs of both sides, we can get the value of Ka which is found to be 7.31
Thus dissociation constant of acetic acid = 7.31
Therefore, equilibrium constant ,Keq = dissociation constant x activity coefficient
Keq = 7.31 x 0.9 = 6.579
Thus, equilibrium constant of the given reaction is 6.579.
Colculate the eghilie riam Constant in Concentrations of the acidi'c disso eackion Cconstant dissociatiort) of acetic acid if the eguilisniun constant is knoun that tle coettisientS of activi...
Acetic acid (CH3COOH) dissociates in water with an equilibrium constant of 1.8x10^-5. Calculate the equilibrium concentrations of all the compounds involved in that process if 2.0M CH3COOH is initially mixed with 1.0M CH3COOH and 1.0M H3O, then allowed to reach equilibrium. Also, determine the pH and pOH of the solution.
What is the ratio of the concentrations of acetate ion and undissociated acetic acid at pH 4.10? (The pKa of acetic acid is 4.76.)
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.10 M buffer solution at pH 4.9? Note that the concentration and/or pH value may differ from that in the first question. 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A^-]/[HA]. 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid. That is, there are...
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.20 M buffer solution at pH 5.0? Note that the concentration, pH value, or both may differ from that in the first question. Strategy 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid). [A-V[HA). 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid. Step 1:...
What concentrations of acetic acid (pKa 4.76) and acetate would be required to prepare a 0.20 M buffer e may differ from that in the first question. STRATEGY 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A]/[HA] 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid. That is, there are 1.7 molecules acetate Step 1: The ratio of base to...
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 4.9? Note that the concentration and/or pH value may differ from that m the first question. STRATEGY 1. Rearrange the Hendereon-Hasselbalch equation to solve for thfe rabo of base (acetate) to aad (acetic add).[A^-]/[HA]. 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration ot acetic acid.
*2.4 Consider a 1M water-solution of acetic acid. a) Given the dissociation constant of acetic acid, K=1.75*10--M, calculate the concentration of non-dissociated acetic acid at equilibrium. Hint The dissociation process is described as a first order process relative to each of the components: CH3COOH CH3COO- + H+
Key Ouestion #2: Determine the initial concentrations for the following solutions. Solution Preparation Acetic Acid Acetate Ion Concentration Concentration Solution #1 50.0 mL of 0.0500 M acetic acid and 0.205 grams of sodium acetate Solution #2 25.0 mL of 0.0500 M acetic acid, 25.0 mL of deionized water, and 0.205 grams of sodium acetate 50.0 mL of 0.0500 M acetic acid and 0.102 grams of sodium acetate Solution #3 Solution # 4 50.0 mL of 0.0500 M acetic acid Solution...
What concentrations of acetic acid (pKa=4.76)(pKa=4.76) and acetate would be required to prepare a 0.10 M0.10 M buffer solution at pH 4.6pH 4.6? Note that the concentration, pH value, or both may differ from that in the first question. Strategy Rearrange the Henderson–Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A−]/[HA][A−]/[HA] . Use the mole fraction of acetate to calculate the concentration of acetate. Calculate the concentration of acetic acid. Step 1: The ratio...
. Calculate the ionization constant for a weak acid, acetic acid (CH3COOH), that is 1.60% ionized in 0.0950 M solution. (Use the ICE table for the dissociation equation) (K. = 1.8 x 10-5 for the acetic acid)