Question

prepare .1M Sodium phospahte buffer, pH 6.8. you need 500ml.

NaH2PO4-H20= 137.99g/mol

Na2HPO4-7H20= 268.08g/mol

Answer 1

We can create phosphate buffer by adding NaH2PO4.H2O and Na2HPO4.7H2O. The buffer action is due to inter conversion of the acid  H2PO₄^- to the salt HPO₄^2- .

H2PO₄^-(aq) + H2O(l) ------> HPO₄^2- (aq) + H3O⁺  , pKa = 7.21

The required pH = 6.8

Applying Hendersen equation

pH = pKa + log [salt] / [acid] = pKa + log [HPO₄^2-(aq)] / [H2PO₄^-(aq)]

=> 6.8 = 7.21 +  log [HPO₄^2-(aq)] / [H2PO₄^-(aq)] ----- (1)

Let's represent  HPO₄^2-(aq) as B and  H2PO₄^-(aq) as A

=> > 6.8 = 7.21 +  log[B] / [A]

=>   log[B] / [A] = 6.8 - 7.21 = - 0.41

=>  [B] / [A]= antilog (- 0.41) = 10^-0.41 = 0.389

=> [B] / [A]  = 0.389 --------- (2)

Since volume (V= 500 mL) is same for both A and B, from eqn(2)

[(moles of B) / V] / [(moles of A) / V] = 0.389

=> (moles of B) / (moles of A) = 0.389

=>  (moles of B) = 0.389x(moles of A) ---------- (3)

We need to Prepare 500 mL (= 0.500 L) of 0.1 M phosphate buffer.

Hence moles of A and B in the phosphate buffer = VxM = 0.500Lx0.1mol/L = 0.0500 mol

Hence   (moles of B) + (moles of A)= 0.0500 mol ------(4)

=>  0.389x(moles of A) + (moles of A) = 0.0500 mol [from eqn(3)]

=> 1.389x(moles of A) = 0.0500 mol

=>(moles of A) = (0.0500 /1.389) mol = 0.0360 mol

Hence moles of B = 0.0500 - (moles of A) = 0.0500 - 0.0360 = 0.014 mol

Mass of A (NaH2PO₄.H2O) = 0.0360 mol x (137.99 g/mol) = 4.97 g NaH2PO₄.H2O

Mass of B(Na2HPO₄.7H2O) = 0.014 mol x (268.08 g/mol) = 3.75 g Na2HPO₄.7H2O

Hence we need to add 4.97 g NaH2PO₄.H2O and 3.75 g Na2HPO₄.7H2O and dilute them to 500 mL to prepare a buffer of pH 6.8 (answer)