Question

A buffer is constructed by dissolving 2.05 grams of sodium benzoate (144.11 g/mol) and 2.47 grams of benzoic acid (122.12 g/mol Ka = 6.50 x 10^-5) in 150.0 mL of distilled water.  The solution is titrated with 0.500 M NaOH. Calculate the pH of the solution:

a) at the equivalence point  

b) 20.00 mL NaOH before the equivalence point  

c) 20.00 mL NaOH after the equivalence point

Answer 1

moles of sodium benzoate, C6H5COONa added = mass / molar mass = 2.05 g / 144.11 g/mol = 0.0142 mol

moles of benzoic acid, C6H5COOH added = mass / molar mass = 2.47 g / 122.12 g/mol = 0.0202 mol

Total volume of the solution, V = 150.0 mL = 0.150 L

Hence [C6H5COONa] = 0.0142 mol / 0.150 L = 0.0947 M

[C6H5COOH] = 0.0202 mol / 0.150 L = 0.135 M

(a): At equivalence point all of the C6H5COOH is neutralized by NaOH to produce C6H5COONa.

Moles of NaOH needed to reach equivalnce point = moles of available C6H5COOH = 0.0202 mol

=> 0.0202 mol NaOH = MxV(L) = 0.500 mol/L x V(L)

=> V(L) = 0.0202 / 0.500 = 0.0404L = 40 mL NaOH

Hence 40 mL of NaOH need to be added to reach equivalence point.

Hence total volume at equivalence point, Vt = 150 mL + 40 mL = 190 mL = 0.190 L

total moles of C6H5COONa at equivalnece point = 0.0142 mol + 0.0202 mol = 0.0344 mol

Hence concentration of C6H5COONa at equivalnece point , [C6H5COONa] = 0.0344 mol / 0.190 L = 0.181 M

Now pH at equivalnece point can be calculated from salt hydrolysis frmulae which is

pH = (1/2)(pKa + pKw + log[C6H5COONa]) = (1/2)x[4.187 + 14 + log(0.181M)] = 8.72 (answer)

(b): 20.00 mL NaOH before the equivalence point :

moles of NaOH added = MxV = 0.500 mol/L x 0.020 L = 0.01 mol

Now 0.01 mol NaOH will react with 0.01 mol C6H5OOH to form 0.01 mol C6H5COONa.

Hence moles of C6H5COONa = 0.0142 + 0.01 = 0.0242 mol

moles of C6H5COOH = 0.0202 - 0.01 = 0.0102 mol

Now applying Hendersen equation

pH = pKa + log[C6H5COONa] / [C6H5COOH] = 4.187 + log(moles of C6H5COONa / moles of C6H5COOH)

=> pH = 4.187 + log(0.0242 mol / 0.0102 mol) = 4.56 (answer)

(c): Moles of excess NaOH added = MxV = 0.500 mol/L x 0.020 L = 0.01 mol = [OH-]

=> pOH = - log[OH-] = - log(0.01M) = 2

=> pH = 14 - 2 = 12 (answer)