Question

The Pka of acrylic acid is 4.25.

What is the concentration of H3O+ in .190M, C3H3O2- in .190M, HC3H3O2 in .19M, OH- in .19M, and the % dissociation in .05M acrylic acid

The pH in .19M of acrylic acid is 2.49

Could someone please show me the steps to this? Thanks

Answer 1

Pho = 4.25 [criven] 2 кои Acrylic acid Lor + H₂O tu 0:19.M. 0 0 Initial Concentration 0.19-X Concentration 2 x at equilibrium pka -log ka [where - logka Constant Edissociation Hon] log s 4.95 = - 4.25 ka Antilog 4.25 ka the il Ka = 5.62 X 10 but ka {p (whese [p] = Concentrattos [R] of products and CR]= Concentration of Reactant) Put value of ta in above equation. 5.62x10-5 = (%) () [0-9-20) on sol solving it, Where Tur] = Cone, ph= I who of 4t gons = x x2 019-X x = 3:24. X10-3 log [ut] pH = -log (x]

PH -log (3.24 X10-3) log / 24x16 3 an solving P4 - 2.49 Conc. of an = conc. of 4t = x 3.24 X/03M M put 0.186.76M Concentratimit acylic acid = {0119-8) at equilibrium the value ofx Conc. of acrylle acid = (6.19 19 -3.24x103) [agon product of [at] [on] =104 = 10-19 put the value of CHt] 3124-X103) Comes ] =10 But Htand on + (3 Сон:] 10-19 Сон-) 3-24103 3.09 *10-12 Percentage of, alsociation of 0.05M quylic acid for this we have recalculate value of x again. to Thue

Page on +40 Inihag cale. 0.05 conc at 0.05 - x 20 equilibrium 옮 ka (2 ] [w 1 0.05-7 calculated above Ka as 5.62 x 10 5 - 2 2 o.os-x Xter Neglecting a solve it 5 162 x 10-5 x2 6o5 x 2 5.62x1050.05 = 0.281_X10-5 2.818106 202. X2 x = 1.676 X103 x Therefor, Cone of disociated Vi dissociation (conc. af dissociated sone Ions. Total concentration x 100 to disociation 10676870 3X 100 = 3: 352 0.05 to disociation = 3.35%.