Question

Describe the preparation of 10 liters of 0.02M phospate buffer, ph 6.9, starting from

(a) a 2M H_​3PO_​4​ solution and a 1 M KOH solution

(b) 1M solutions of KH_​2PO_​4, and Na₂HPO₄

_​​(c) solid Na₃PO₄​ and 1M HCl

Please provide detailed explanations!

Answer 1

Given data Concentration of phosphate buffer Volume of phosphate buffer-10 L pH of phosphate buffer, pH 6.9 0.02 M Calculate the moles of phosphate buffer(n Tphosghate buffer Concentration of phosphate buffer x Volume of phosphate buffer Substitute the values in the above expression phoshae buffer 0.02x10 0.2 mol Given data Concentration of HPO HPO2 M Concentration of KOH, [KOH 1M Consider the chemical equation pk, the above reaction is 7.2. Write the expression to calculate the volume of H3PO4 ( ,PO hosprate buffer Substitute the values in the above expression 0.2 HjPO*2

Therefore, the volume of HPO4 is 100 mlL Consider the expression of the pH of buffer solution HPO- pH = pk, +10g Substitute the values in the above expression 6.98 7.2 tle LHPO:1 H.PO:] H, -6.9-7.2 HPO2 los H,PO: -0.3 HPO H,PO: 0:01 [HPO) = 0.501 H.PO.] Therefore, Here Solve equation (1) and equation (2) H,PO4-0133 mol HP0"一0.067 mol Consider the expression to calculate the volume of KOH needed to make HPO Vmoles of HPO2 кон Substitute the values in the above expression. 0.067 KCH = 0.067 L - 67 mL

Given data Concentration of [KH.Po, ] + [Na,HPO,] = 1 M Consider the chemical equation of dissociation. pk, the above reaction is 7.2. Consider the expression of the pH of buffer solution HPO2-7 pH = pk, +10g Substitute the values in the above expression. HPO2 6.9 = 7.2 + log [HPOİ ] [H,Po69-12 HPO lo 40.3 HPO H.POİ = 0.501