CaCO3 (s) scaz-(aq)\"CO32-(aq) The solubility of CaCO3 at 25°C is 690-10-5 M. Th
ID: 1027723 • Letter: C
Question
CaCO3 (s) scaz-(aq)"CO32-(aq) The solubility of CaCO3 at 25°C is 690-10-5 M. The reaction is: a) Calculate the concentration of Ca2 and CO32 at equilibrium v. solubility product.) ylD (Answer) b) Calculate the equilibrium constant for the dissolution reaction. (This equilibrium constant is also known as the vi. A 0400 M Hcio solution was found to have an H+ concentration of 1 10 io 4 M The reaction HOO aq)-mog) a) Calculate the equilibrium constant for the ionization reaction (also known as the acid dissociation constant, o) the acid dissociation constant, y (-(A (A b) Calculate the ratio ( (HCIO)/ [HOO] n tial). 100 This value is also known as the percent ionization (al vi. At 1285 C the equilibrium constant for the reaction: Br2g) 2Br(g) is Ke104-10 A 0.200-L vessel containing an equilibrium mixture of the gases has 0.245 g Br2(g) What is the mass of Br(g) in the vessel? Answer) vili. For the reaction: Co2(g) H2(g) COg) H2(g), the equilibrium concentrations are [CO2] 00190 M, [H21 0.0340 M, [Co] [H20] 0.0220 M at a temperature of 298 K when contained in a volume of 1.00L a) What is the value of Ke? (Answer) b) What is the value of Kp? Answer) b) What are the concentrs of the four substances n le fraction? xco (Answer) c) How does the value of Kc compare if the k is determined using mole fraction instead of molarity,KC (M) ? (Answer in words and numbers)Explanation / Answer
V)the equilibrium expression for reaction is as follows.
Ksp =[Ca+2]CO32-]
IN THE REACTION WE ALWAYS WRITE SPARINGLY SOLUBLE SUBSTANCE ,SOLID ON REACTANT SIDE AND DISSOLVED IONS ON PRODUCT SIDE.
Ksp is known as solubility product.solid doesn’t appear in the equilibrium.
Solubility refers to moles of solid that dissolve in liter of solution.For every one molecule of solid calicium carbonate dissolves we get one calcium ion and one carbonate ion.the solubility expression for these ions is
S = [Ca+2] . Or S =[CO3]2-
THE CONCENTRATION OF Ca+2= solubility of CaCO3=6.9 X 10-5M
CONCENTRATION OF CO32- =6.9 X 10-5M
SUBSTITUTE SOLUBILITY EXPRESSIONS IN Ksp EXPRESSION.
Ksp =(S)(S)=S2
Ksp =(6.9 x10-5)2 = 47.61 x10-10
vi) Ka = [H+][ClO-]/ [HClO]eq
[H+]eq=[ClO-]eq ,[HClo]eq ={HClOInitial]-x(change) =0.4 -0.00011 =0.39989
Ka=(1.1 x10-4)2/ 0.39989 =3.025 x10-8
B)percent ionization = ([H+]/[HClO]initial)*100 (WE USE H+ CONCENTRATION IN NUMERATOR BECAUSE THIS IS THE AMOUNT OF ACID THAT EXIST AS IONS AT PARTICULAR CONCENTRATION)
PERCENT IONISATION = (1.1 X10-4/0.4)*100
PERCENT IONIZATION =0.0275 %