The beakers shown contain 0300 L of aqueous solutions of a m

The beakers shown contain 0.300 L of aqueous solutions of a moderately weak acid HY. Each particle represents 0.0200 mol; solvent molecules are omitted for clarity. Protons are smaller and light blue. Oxygen atoms are red. Therefore, hydronium ions are red with three light blue hydrogen atoms Hint Solution Guided Solution points eBook (a) The reaction in beaker A is at equilibrium. Caleulate Q for B, C, and D to determine which, if any, are also at equilibrium. Print 0c\" 2D- O Bisat equilibrium C is at equilibrium. D is at equilibrium. O (b) For any not at equilibrium, in which direction does the reaction proceed? B proceeds to the right. Bproceeds to the left. C proceeds to the right. C proceeds to the left. D proceeds to the right. Dproceeds to the left.

Solution

a) HY +H2O <--->H3O+ + Y-

note: # of red-blue=# of H3O+=# y-(green only)

(# of greens+ green blue) - (# of red -blue) = # of HF

At equilibrium, [Y-]=[H3O+]=4*0.02mol=0.08mol

# of HY(green -blue)=total (green+green blue) -(red-blue)=11-4=7

[HY]=0.02*7=0.14mol

Keq=[Y-][H3O+]/[HY]=(0.08mol)(0.08mol)/0.14mol=0.04

equilibrium constant=0.04

For B,

# of red-blue=2=# of H3O+=# of Y-

mol of H3O+ and Y-=2*0.02=0.04mol

# of HY=7-2=5

mol of HY=5*0.02=0.1

QB=[H3O+][Y-]/[HY]=(0.04)(0.04)/(0.1)=0.016

For C,

# of red-blue=2=# of H3O+=# of Y-

mol of H3O+ and Y-=2*0.02=0.04mol

# of HY=6-2=4

mol of HY=4*0.02=0.08

QC=[H3O+][Y-]/[HY]=(0.04)(0.04)/(0.08)=0.02

For D,

# of red-blue=2=# of H3O+=# of Y-

mol of H3O+ and Y-=2*0.02=0.04mol

# of HY=4-2=2

mol of HY=2*0.02=0.04

QD=[H3O+][Y-]/[HY]=(0.04)(0.04)/(0.04)=0.04=Keq

Thus QD =Keq so D is at equilibrium

b) QC=0.02<Keq=0.04 ,so C proceeds to the left

QB=0.016<Keq=0.04 ,so B proceeds to the left