Question

The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a solution of nac6h5coo if the ph is 9.04?

Answer 1

$NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}$

Here the base is a benzoate ion, which is a weak base and reacts with water.

$C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)$

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 $[OH^{-}] = 10^{^{-pOH}}$

$[OH^{-}] = 10^{^{-4.96}}$

$[OH^{-}] = 1.1\times 10^{-5}$

The base dissociation equation kb = $\frac{Product}{Reactant}$

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

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 = $1.6\times 10^{-10}$

And value of [OH-] we have calculated as $1.1\times 10^{-5}$ and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

$1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}$

$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

$[C6H5COO^{-}] = 0.76 M$ or $0.76\frac{mol}{L}$

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 = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

Answer 2

Answer : The number of moles of sodium benzoate is, 0.375 moles

Solution :

First we have to calculate the pOH.

$pOH=14-pH=14-9.04=4.96$

Now we have to calculate the concentration of [tex[OH^-[/tex] ions.

$pOH=-\log [OH^-]$

$4.96=-\log [OH^-]$

$[OH^-]=1.096\times 10^{-5}$

The equilibrium reaction for dissociation of weak base is,

                   $C_6H_5COO^-+H_2O\rightleftharpoons OH^-+C_6H_5COOH$

initially conc.         c                              0         0

At eqm.              $c(1-\alpha)$                      $c\alpha$        $c\alpha$

The expression for dissociation constant is,

$k_b=\frac{c\alpha\times c\alpha}{c(1-\alpha)}$

when $\alpha$ is very very small the, the expression will be,

$k_b=\frac{c^2\alpha^2}{c}=c\alpha^2\\\\\alpha=\sqrt{\frac{k_b}{c}}$

And,

$[OH^-]=c\alpha$

Thus the expression will be,

$[OH^-]=\sqrt{k_b\times c}$

Now put all the given values in this expression, we get

$1.096\times 10^{-5}=\sqrt{(1.6\times 10^{-10})\times c}$

$c=0.75M$

Now we have to calculate the moles of sodium benzoate.

$Moles=concentration\times volume=0.75M\times 0.5L=0.375moles$

Therefore, the number of moles of sodium benzoate is, 0.375 moles