Question

What is the pH of the solution using the henderson-hasselbach equation for an aqueous solution in a 55-gallon (208L) drum, characterized by minimal buffering capacity, contains 6 kg of phenol and 3 kg of sodium phonate.

Answer 1

According to Henderson-Hasselbach equation,

pH= pK_a + log [salt]/[acid],

here the pK_a of phenol is 10.0.

no.of moles of phenol = 6/(.094) = 63.8 mol where the molecular weight of phenol is 94.0 g/ mol.

no.of moles of sodium phenolate = 3/(.116) = 25.86 mol where the molecular weight of sodium phenolate is 116.0 g/ mol.

now, 208 L of water contains 63.8 mol of phenol,

Hence, 1 L of water will contain 63.8/208 * 1 =0.238 mol of phenol,

Similarly,

208 L of water contains 25.86 mol of sodium phenolate,

Hence, 1 L of water will contain25.86/208 * 1 =0.124 mol of sodium phenolate.

So the [salt] = 0.124,

[acid]=0.238

Substituting in the Henderson-Hasselbach equation we get,

pH= 10 + log (0.124 / 0.238)

= 9.716