Question #70bb0

pH of 0.2M #HC_7H_5O_2#

notice: #H^(+)# is really #H_3O^(+)# just abbreviated.

Since our reaction would look like this:
#HC_7H_5O_2#(aq)#+H_2O#(l)#rightleftharpoonsH^(+)#(aq)#+C_7H_5O_2^(-)#(aq)

ICE TABLES
a)#HC_7H_5O_2#
I) 0.2M
C) -x
E) #0.2M-x~~0.2M# This only works with weak acids/bases, because the x you solve for will be extremely small.

b)#H^(+)#
I) 0
C) +x
E) #0+x=x#

c)#C_7H_5O_2^(-)#
I) 0
C) +x
E) #0+x=x#

The #Ka=([Products])/([Reactants])##=([H^(+)]*[C_7H_5O_2^(-)])/([HC_7H_5O_2])##=(x*x)/(0.2M)=x^2/(0.2M)#

Notice: a) #[X]=#Molar Concentration of X
b) We leave #H_2O# out of our Ka because its concentration is constant. This happens with any pure liquids or solids. If gases in equation, use partial pressures.

Since #Ka=6.3*10^-6##=x^2/(0.2M)#

#x=sqrt((6.3*10^-6)*(0.2M))##=1.12*10^-3M=[H^(+)]#

NOW #pH=-log[H^(+)]=-log1.12*10^-3M#

#pH=2.94#

Extra step for weak bases: If you are dealing w/ a weak base and your #x=[OH^(-)]#, this last step will give you #pOH#
(#pOH=-log[OH^(-)]#)

You can use #pOH# to solve for #pH# with this simple relationship:
#pOH+pH=pKw# since #pKw=14.00#

#pH=14.00-pOH#

1. #"pH"=2.95#

2. #"pH"=9.67#

A quick way is to use:

#pH=1/2(pK_a-logA)#

#A# = conc. of the acid.

#pK_a=-logK_a=-log(6.3xx10^(-6))=5.2#

#pH=1/2(5.2-log0.2)=1/2(5.2+0.698)= color(red)2.95#

We can use the same idea for weak bases:

#pOH=1/2(pK_b-logB)#

Where #B# = conc. of the base.

#pK_b=-logK_b=-log(1.1xx10^(-8))=7.958#

#pOH=1/2(7.958-log0.2)=1/2(7.958+0.698)=4.33#

#pH+pOH=pK_w=14#

#pH=14-4.33=color(red)9.67#