Question #18488

1 Answer
Jan 22, 2017

Answer:

The degree of dissociation #sf(alpha=0.0158)#

#sf(K_b=2.51xx10^(-6)color(white)(x)"mol/l")#

Explanation:

Triethyamine is a weak base and ionises:

#sf((CH_3)_3N+H_2Orightleftharpoons(CH_3)_3stackrel(+)(N)H+OH^-)#

For which:

#sf(K_b=([(CH_3)_3stackrel(+)(N)H][OH^(-)])/([(CH_3)_3N]))#

Rearranging and taking -ve logs of both sides we get the expression:

#sf(pOH=1/2[pK_b-logb])#

#sf(b)# is the concentration of the base at equilibrium. Because the dissociation is small we can assume that the initial concentration is a good enough approximation to that at equilibrium.

We can get #sf(pOH)# using the expression:

#sf(pH+pOH=14)# at 298K

#:.##sf(pOH=14-pH=14-10.2=3.8)#

#:.##sf(3.8=1/2[pK_b-log(0.01)]=1/2pK_b+1)#

#:.##sf(1/2pK_b=2.8)#

#sf(pK_b=5.6)#

#:.##sf(-logK_b=5.6)#

#:.##sf(K_b=10^(-5.6)=2.51xx10^(-6)color(white)(x)"mol/l")#

The degree of dissociation #alpha# is the fraction of the base which has dissociated.

We can find the concentration of #sf(OH^-)#:

#sf(pOH=3.8=-log[OH^-])#

#:.##sf([OH^-]=10^(-3.8)=1.585xx10^(-4)color(white)(x)"mol/l")#

#sf(alpha=[[OH^-]]/[[(CH_3)_3N]])#

#:.##sf(alpha=1.585xx10^(-4)/(0.01)=0.0158)#

The standard, if not more long - winded method, is to set up an ICE table in concentrations (M):

#" "sf((CH_3)_3N+H_2Orightleftharpoons(CH_3)_3stackrel(+)(N)H+OH^-)#

#sf(I" "c" "0" "0)#

#sf(C" "-calpha" "+calpha" " +calpha)#

#sf(E" "c(1-alpha)" "calpha" "calpha)#

#:.##sf(K_b=(c^cancel(2)alpha^(2))/(cancel(c)(1-alpha))=(calpha^2)/((1-alpha))#

Because the dissociation is small we assume that #sf((1-alpha)rArr1)#

#:.##sf(K_b=calpha^2)#

#:.##sf(alpha^2=K_b/c)#

#:.##sf(alpha=sqrt(K_b/c))#

#sf(alpha=sqrt((2.511xx10^(-6))/0.01)#

#sf(alpha=0.0158)#

Whichever method you prefer gets you the same answer.