Respuesta :
[tex] NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-} [/tex]
Here the base is a benzoate ion, which is a weak base and reacts with water.
[tex] C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq) [/tex]
The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.
Therefore [OH-] = [C6H5COOH]
In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]
pOH = 14 - pH
pH given = 9.04
pOH = 14-9.04 = 4.96
pOH = -log[OH-] or [tex] [OH^{-}] = 10^{^{-pOH}} [/tex]
[tex] [OH^{-}] = 10^{^{-4.96}} [/tex]
[tex] [OH^{-}] = 1.1\times 10^{-5} [/tex]
The base dissociation equation kb = [tex] \frac{Product}{Reactant} [/tex]
[tex] kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}[/tex]
H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.
Value of Kb is given = [tex] 1.6\times 10^{-10}[/tex]
And value of [OH-] we have calculated as [tex] 1.1\times 10^{-5} [/tex] and value of C6H5COOH is equal to OH-
Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-
[tex] kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}[/tex]
[tex] 1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]} [/tex]
[tex] [C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}} [/tex]
[tex] [C6H5COO^{-}] = 0.76 M [/tex] or [tex] 0.76\frac{mol}{L} [/tex]
So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L
Moles of NaC6H5COO would be = [tex] 0.76(\frac{mol}{L}) \times (0.50L) [/tex]
Moles of NaC6H5COO (sodium benzoate) = 0.38 mol
Answer : The number of moles of sodium benzoate is, 0.375 moles
Solution :
First we have to calculate the pOH.
[tex]pOH=14-pH=14-9.04=4.96[/tex]
Now we have to calculate the concentration of [tex[OH^-[/tex] ions.
[tex]pOH=-\log [OH^-][/tex]
[tex]4.96=-\log [OH^-][/tex]
[tex][OH^-]=1.096\times 10^{-5}[/tex]
The equilibrium reaction for dissociation of weak base is,
[tex]C_6H_5COO^-+H_2O\rightleftharpoons OH^-+C_6H_5COOH[/tex]
initially conc. c 0 0
At eqm. [tex]c(1-\alpha)[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
The expression for dissociation constant is,
[tex]k_b=\frac{c\alpha\times c\alpha}{c(1-\alpha)}[/tex]
when [tex]\alpha[/tex] is very very small the, the expression will be,
[tex]k_b=\frac{c^2\alpha^2}{c}=c\alpha^2\\\\\alpha=\sqrt{\frac{k_b}{c}}[/tex]
And,
[tex][OH^-]=c\alpha[/tex]
Thus the expression will be,
[tex][OH^-]=\sqrt{k_b\times c}[/tex]
Now put all the given values in this expression, we get
[tex]1.096\times 10^{-5}=\sqrt{(1.6\times 10^{-10})\times c}[/tex]
[tex]c=0.75M[/tex]
Now we have to calculate the moles of sodium benzoate.
[tex]Moles=concentration\times volume=0.75M\times 0.5L=0.375moles[/tex]
Therefore, the number of moles of sodium benzoate is, 0.375 moles
