Chemistry Help, Determining Ka/Kb?
you are incorrect the Ka expression would be Ka= [C2H3O2][H3O+]/[HC2H3O2] Water is neglected because it is not aqueous it is a liquid and therefore has no concentration. H3O+ is the same thing as H+ so the concentration of H+ is H3O+ concentration as well The Ka value would be Ka= (1.27x10^-4)(1.27x10^-4) / (0.00100 M) The reason A- is the same as H+ is because of the equation HA <===> H+ + A- every mole of acid, HA, produces one mole of H+ and one mole of A- there their concentrations must be the same as well.
In chemistry, what is " Kw"?
Water can act as both an acid or base, and water molecules dissociate into hydronium (acidic) and hydroxide (basic) ions, which react with each other to form water again! Thus, these ions are in equilibrium with the water molecule. This is called auto ionization of water.(Source of image: Ch_15 - Acids and Bases)At any point of time, very little amount of the water molecules are actually ionized, so we assume that the concentration of water (in liquid form) to be virtually unchanged.Now, the equilibrium constant expression for any reaction isK= [Products]/[Reactants]Since the concentration of water remains constant relative to the ion’s concentrations, it is not included in the equilibrium constant expression for the above reaction. Thus, the equilibrium constant (Kw) expression for the auto ionization of water is:Kw= [H3O+][OH-]Now, the equilibrium constant of any reaction is a constant, and it only varies with the temperature (and other parameters like solvent, etc.). Since the concentration of the hydronium ions will also change if Kw changes, it makes sense that the pH of water changes at different temperatures.pH=−log[H3O+](Source of image: http://slideplayer.com/slide/752...)One can see that the Kw remains of the order 10E(-14) at a wide range of temperatures, and therefore, for most calculations, it is considered as 10E(-14).
Chemistry Help: pH & Kb?
Ka for acetic acid = 1.8 x 10^-5 1.8 x 10^-5 = x^2 / 0.079-x x = [H+]= 0.0012 M % dissociation = 0.0012 x 100/ 0.079=1.5 Ka for HF = 6.8 x 10^-4 Kb for F- = Kw/Ka = 1.0 x 10^-14 / 6.8 x 10^-4 =1.5 x 10^-11 Ka for phenol = 1.6 x 10^-10 Kb for phenolate C6H5O- = Kw/Ka = 1.0 x 10^-14 / 1.6 x 10^-10 =6.3 x 10^-5 C6H5O- + H2O < => C6H5OH + OH- 6.3 x 10^-5 =x^2 / 0.140-x x = [OH-]= 0.0030 pOH = - log 0.0030=2.5 pH = 14 - 2.5 = 11.5 Ka for CH3NH3+ = Kw/Kb = 1.0 x 10^-14 / 4.4 x 10^-4 =2.3 x 10^-11 CH3NH3+ <=> CH3NH2 + H+ 2.3 x 10^-11 = x^2 / 0.19-x x = [H+]= 2.1 x 10^-6 M pH = 5.7
Chemistry help on pH and Kb/Ka?
1). pOH = pKb + log_10 [conjugate acid]/[weak base] pOH = -log_10 (8.2 10^-5) + log_10 (0.580 / 0.180) = 4.59 pH = 14 - 4.59 = 9.41 2). pH = pKa + log_10 [conjugate base]/[weak acid] pH = -log_10 (8.4 10^-5) + log_10 (0.450 / 0.100) = 4.73 . ..
Ka and Kb Chemistry problem.?
Acetic acid is the conjugate acid of the acetate anion. It is a weak acid that dissociates to an acetate ion and a hydronium ion in aqueous solution. Calculate Ka for acetic acid if a 1.0M solution results in an equilibrium H3O^+ concentration of 0.0042M
Chemistry, Ka Kb for conjugate base.?
Equation: Ka X Kb = 10^-14 If Ka = 4.9*10-10 Then Kb= 10^-14 / 4.9*10^-10 Kb = 2.04*10^-5
Chemistry help needed! (calculating the Ka or Kb value of salt in solution) *help?*?
Sodium acetate dissolves totally in water. Some of the acetate ion reacts with water: OAc- + H2O = HOAc + OH- Since the pH + pOH = 14, pOH= 4.75. Then OH- = 10^-4.75 = 10-5 * 10^(.25) = 1.8x10-5 Since [OH]- is of the order of 10-5 and equals [HOAc], compared to an initial conc of 1 M of acetate ion, the equilibrium concentration is just about 1 M. Define Kb= [HOAC][OH-]/[OAc]-, then, We get Kb = [1.8x10-5]^2 = 3.2 x 10-10. BTW, Kb*Ka = Kw, as noted by initial answerer. Since you asked for either Kb or Ka, this approach was simpler.
Chemistry Help! Ka/Kb acid base hydrolysis?
First , calculate the molarity of acetic acid : 30.0 g acetic acid ( 1 mole / 60 g ) = 0.5 mole = 0.5 M Now , you have a solution of 0.5 M acetic acid with pKa of 4.74 : Let x be the equilibrium concentration of H+ ion . Form an ICE table : CH3COOH + H2O ----> CH3COO- + H3O+ ------- C ------------------------- 0 ------------- 0 ------- C-x ---------------------- x ------------- x Substitute x in the constant of the equation : Ka = (x)(x)/C-x Ka = x^2/C-x Solve the equation for x and you'll get x=[H+]= 3×10^-3 pH = -log [H+] pH = -log (3×10^-3) pH = 2.52 Good luck ;)
Physical Chemistry: What is the difference between Ka, Kb, and Kw?
Ka is acid equilibrium constant, Kb is equilibrium constant when a base and its conjugate acid are in equilibrium. Now where as Kw is totally different equilibrium constant known as ionic product of water. Kw is specific to water(or H2O) in equilibrium with H+ and OH- ions. I gave what ever I understood being an MSc and M.Phil in chemistry in central University
Chemistry calculate pH given Kb or Ka and molarity. Please help me figure out how to solve these!!!?
a) pOH =0.5(pKb - lg(0.5)) = 1.78 => pH=12.2 b) pH =0.5(pKa - lg(0.79)) = 1.62