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geniusboy [140]
3 years ago
12

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 soluti

on of nac6h5coo if the ph is 9.04?
Chemistry
2 answers:
lyudmila [28]3 years ago
8 0

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

mr_godi [17]3 years ago
5 0

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

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