For the following set of questions, consider an aqueous solution of 0.0180-F of
ID: 989148 • Letter: F
Question
For the following set of questions, consider an aqueous solution of 0.0180-F of benzoic acid (K_a = 6.28 times 10^-5). Write the charge balance AND mass balance expressions for the above solution. Calculate the equilibrium concentration of H^* in the solution. List two assumptions that are made in the previous calculation Consider the above solution with the individual addition of the salts below (i.e. four different solutions with one extra salt each). Order these solutions by increasing P^H (i.e. from lowest to highest). 0.10 F NaCl, 0.15 F KCl 0.03 F Li_3,PO_4 0.05 F Na_2So_4Explanation / Answer
7
a)
Assume HB = Benzoic Acid, B is the benzoate ion and H+ the acidic pproton
C7H5O2H --> benzoic acid
C7H5O2- -->benzoate ion
H+ --> hydrogen ion (acidic)
then
C7H5O2H + H2O <--> C7H5O2- + H3O+
this is the ionization of the weak acid in water
then
Ka = [C7H5O2-][H3O+]/[C7H5O2H ]
Ka = 6.28*10^-5
Charge balance is essentially
C7H5O2H + H2O <--> C7H5O2- + H3O+
left side: all neutral i.e. = 0
righ side: -1 +1 = 0
equation is balanced
b)
find H+ in equilibrium
since
Ka = [C7H5O2-][H3O+]/[C7H5O2H ]
then
assume
[C7H5O2-] = [H3O+] = x
[C7H5O2H ] = 0.018-x
6.28*10^-5 = (x*x)/( 0.018-x)
solve for x
x = 0.00103
[H+] = x= 0.00103 F
c)
Assumption:
- this is done at 25°C; where Ka and Kw of water are valid
- We assume [H+] = [Benzoate ion] due to stochiometry
d)
consider acid + salt
are salts are neutral, since they are made from strong acids + stron bases
therefore, we need to account for:
total "ionic strenght"
that is
0.1 of NaCl = 0.1 Na + and 0.1 Cl- = 0.20
0.15 of KCl= 0.15 K+ and 0.15 Cl- = 0.30
0.03 of Li3PO4 = 0.09 Li+ and 0.03 PO4-2 = 0.12
0.05 of Na2sO4= 0.10 Na+ and 0.05 PO4-2 = 0.15
then
KCl > NaCl > Na2SO4 > Li3PO4
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