Question

1. 1.22 gm af benzoic acid were dissolved per litre at 25°C (the ionization constant benzoic acid. Ka=6.4 x 10-5) Calculate the pH of the solution?​

Answer 1

Answer:

K[PhCOO−][Ag+]=s×s=s2=2.5×10−13

s=5.0×10−7M. This is solubility of silver benzoate in pure water.

pH=−log[H+]=3.19

[H+]=6.456×10−4

Benzoate ions combine with protons to form benzoic acid but hydrogen ion concentration is constant due to buffer solution.

PhCOOH⇌PhCOO−+H+     

Ka=[PhCOOH][PhCOO−][H+]

[PhCOO−][PhCOOH]=Ka[H+]=6.46×10−56.456×10−4=10

Let y M be the soubiity in the buffer solution.

y=[Ag+]=[PhCOO−]+[PhCOOH]=[PhCOO−]+10[PhCOO−]=11[PhCO

Thus, silver benzoate is 3.32 times more soluble in buffer

Explanation:

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Answer 2

Explanation:

Answer

Correct option is

B

3.32

Since pH=3.19

[H 3 O + ]=6.46×10 −4 MC 6

H 5 COOH+H 2 O⇋C 6 H 5

COO − +H 3 OK a = [C 6 H 5 COOH]

[C 6 H 5 COO − ][H 3 O + ]

[C 6 H 5 COO − ][C H5 COOH]

= K a

[H 3 O + ]

= 6.46×10 −5 6.46×10 −4

=10

Let the solubility of C 6

H 5 COOAg be x mol/LThen,[Ag + ]=x

[C 6 H 5 COOH]+[C 6 H

5 COO− ]=x10[C6 H 5 COO − ]+[C 6

5 / COO - ]=x[C 6 H 5 COO -- ]= 11xK sp

[AG + ][C 6 H 5

COO _ ]2.5×10 −13

=x( 11x )

x=1.66×10

−6

mol/L

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66×10 −6

mol/L

Now, let the solubility of C 6 H

COOAg be x M

Then, [Ag + ]=x

′

M and [C 6 H 5 COO - ]=x ′

MK sp

=[Ag + ][C 6

H 5 COO - ]K sp

=(x ' ) x ' = K sp

= 2.5×10 −13

=5×10 −7 mol/L

∴ x ,x

= 5×10 −7

1.66×10 −6

=3.32

Hence, C

6 H 5

COOAg is approximately 3.32 times more soluble in low pH solution.

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