Benzoic acid, C_6H_5COOH, has K_a = 6.4 times 10^-5. Its conjugate base, benzoat

Question

Benzoic acid, C_6H_5COOH, has K_a = 6.4 times 10^-5. Its conjugate base, benzoate, is commonly added to food products in small amounts as a preservative. Although the benzoic acid form is more active as a preservative, the benzoate form is significantly more soluble in water and hence easier to handle. In a particular product, it is found that 63% of the added sodium benzoate exists instead in the benzoic acid form. What is the pi I of that material? What percentage of the sodium benzoate would exist in the active benzoic acid form at pH 4.80?

Explanation / Answer

ka of benzoic acid = 6.4*10^-5

pKa of benzoic acid = -log(6.4*10^-5) = 4.20

Let initial mass of benzoic acid = 100 g

.Moles of benzoic acid = 100 g / 122.1 g/mol = 0.82 mol

Moles of benzoate = 63 g / 114 g/mol = 0.552 mol

We know that ,

pH = pKa + log (salt/acid)

= 4.20 + log (0.552/(0.82-0.552))

= 4.51

Now pH = 4.80;

pH = pKa + log (salt/acid)

4.80 = 4.20 + log (salt/acid)

  (salt/acid) = 3.98

x/1-x = 3.98

x = 0.80

So 80 of benzoic acid is in form of benzoate

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