Question

Find the pH of each mixture of acids. Acid Ionization Constants (Ka) for Some Monoprotic Weak Acids at 25 ∘C Acid Formula Ka Benzoic acid HC7H5O2 6.5×10−5 Hydrofluoric acid HF 6.8×10−4 Phenol HC6H5O 1.3×10−10 Formic acid HCHO2 1.8×10−4 Hypochlorous acid HClO 2.9×10−8

Answer 1

Find the pH of each mixture of acids. Acid Ionization Constants (Ka) for Some Monoprotic Weak Acids at 25 ∘C Acid Formula Ka Benzoic acid HC7H5O2 6.5×10−5 Hydrofluoric acid HF 6.8×10−4 Phenol HC6H5O 1.3×10−10 Formic acid HCHO2 1.8×10−4 Hypochlorous acid HClO 2.9×10−8

A.) 0.070 M in HNO3 and 0.185 M in HC7H5O2 Express your answer to two decimal places.

B.) 0.025 M in HBr and 0.020 M in HClO4 '

C.) 0.095 M in HF and 0.230 M in HC6H5O

Answer:

a) the pH of the mixture = 1.13

b) the pH of this mixture  = 1.35

c) the pH of this mixture  = 2.09

Explanation:

The formula to determine the hydrogen ion concentration in benzoic acid is as follows:

$[H^+] = \sqrt{K_2C}$

$[H^+] = \sqrt{6.5*10^{-5}*0.185M}$

$[H^+] =0.0035M$

HNO₃ is a strong acid whose dissociation is complete and thorough into its ions.

$HNO_3 \rightleftharpoons H^+ + NO^-_3$

So, $[H^+]$ = 0.070 M

The total hydrogen concentration of the mixture will now be:

$[H^+]$ = (0.0035 + 0.070 )M

$[H^+]$ = 0.0735 M

Now the pH can be determined as:

pH = - log $[H^+]$

pH = - log [0.0735]

pH = - (-1.13)

pH = 1.13

Hence, the pH of the mixture = 1.13

b) In HBr and HClO₄, these are both strong acids and as such, they undergo complete dissociation of their ions.

$HBr \rightleftharpoons H^+ + Br^-$

So, $[H^+] = 0.025 M$

$HClO_4 \rightleftharpoons H^+ +ClO^-_4$

So, $[H^+] = 0.020 M$

The total hydrogen ion concentration of the mixture =

$[H^+] = 0.020 M + 0.025 M$$[H^+] = 0.045 M$

Now; the pH will be

pH = - log $[H^+]$

pH = - log [0.045]

pH = - (-1.35)

pH = 1.35

Thus, the pH of this mixture  = 1.35

c) For hydrofluoric acid and phenol (weak acids ), the following expression is used to calculate the hydrogen ion concentration

$[H^+] = \sqrt{K_{a1}C_1+K_{a2}C_2}$

$[H^+] = \sqrt{6.8*10^{-4}*0.095 +1,3*10^{-10}*0.230$

$[H^+] = 0.008037 M$

The pH can then be calculated as follows:

pH = - log $[H^+]$

pH = - log  [0.008037]

pH = - (-2.09)

pH = 2.09

Thus, the pH of this mixture  = 2.09

Answer 2

The Ka shows the extent of dissociation of an acid in solution.

For benzoic acid;

HC7H5O2(aq) ⇄ H^+(aq) + C7H5O2^-(aq)

Ka = [ H^+] [C7H5O2^-]

But  [ H^+]  = [C7H5O2^-]

Hence, Ka =  [ H^+]^2

 [ H^+] = √Ka

 [ H^+] = √6.5×10−5

 [ H^+] = 8.1 ×10−3 M

pH = -log [ H^+]

pH = -log[8.1 ×10−3 M]

pH =  2.1

For Hydrofluoric acid;

HF(aq) ⇄ H^+(aq) + F^-(aq)

Ka = [ H^+] [ F^-]

But [ H^+] = [ F^-]

Ka =  [ H^+] ^2

[ H^+] = √Ka

[ H^+] = √6.8×10−4

[ H^+] = 0.0261 M

pH = -log [ H^+]

pH = -log[ 0.0261 M]

pH = 1.58

For Phenol;

HC6H5O(aq)  ⇄ H^+(aq) + C6H5O^-(aq)

Ka = [H^+] [C6H5O^-]

But  [H^+] = [C6H5O^-]

Ka =  [H^+] ^2

[H^+] = √Ka

[H^+] = √ 1.3×10−10

[H^+] =1.14 ×10−5 M

pH = -log [H^+]

pH = -log[1.14 ×10−5 M]

pH = 4.94

For Formic acid

HCHO2(aq)  ⇄ H^+(aq) + CHO2^-(aq)

Ka = [ H^+] [CHO2^-]

But  [ H^+] = [CHO2^-]

Ka =  [ H^+] ^2

[ H^+] = √Ka

[ H^+] = √1.8×10−4

[ H^+] = 0.0136 M

pH = - log [ H^+]

pH = -log[ 0.0136 M]

pH = 1.87

For Hypochlorous acid

HClO(aq) ⇄ H^+(aq) + ClO^-(aq)

Ka = [H^+] [ClO^-]

But  [H^+]= [ClO^-]

Ka =  [H^+]^2

[H^+] = √Ka

[H^+] = √2.9×10−8

[H^+] = 1.7 ×10−4 M

pH= -log [H^+]

pH = -log[1.7 ×10−4 M]

pH = 3.77

Learn more: https://brainly.com/question/4147359