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Calculate the pH of a 0.45M solution of Benzonic Acid. (Use the quadratic equation to solve).

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Chemistry

Calculate the pH of a 0.45M solution of Benzonic Acid. (Use thequadratic equation to solve).

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3.6 Ratings (596 Votes)

Answer – Given, [C6H5COOH] = 0.45 M , Ka for the C6H5COOH = 6.4*10-5

C6H5COOH + H2O ------> H3O+ + C6H5COO-

I   0.45                               0           0

C    -x                                 +x          +x

E   0.45-x                            +x          +x

Ka = [H3O+] [C6H5COO-] / [C6H5COOH]

6.4*10-5 = x*x /(0.45-x)

6.4*10-5 *(0.45-x) = x2

Now we need to set up quadratic equation

2.88*10-5 - 6.4*10-5 x = x2

x2 + 6.4*10-5 x - 2.88*10-5 = 0

a =1 , b = 6.4*10-5, c = -2.88*10-5

Using the quadratic equation

x = -b +/- √b2-4a*c / 2a

Plugging the value in this formula

x = 0.00533 M

so, x = [H3O+] = 0.00533 M

so, pH = -log [H3O+]

           = -log 0.00533 M

           = 2.27

So, the pH of a 0.45M solution of Benzonic Acid is 2.27
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