Preparing Buffers! Molecular weights: NaH2PO4: 119.96 g/mol Na2HPO4: 141.96 g/mo
ID: 883465 • Letter: P
Question
Preparing Buffers!
Molecular weights:
NaH2PO4: 119.96 g/mol
Na2HPO4: 141.96 g/mol
Tris base: 121.14 g/mol
Tris-HCl: 157.60 g/mol
Sodium acetate: 58.04 g/mol
pKa:
Tris: 8.30
Phosphate: 7.21
Acetate: 4.76
1) Prepare 750 mL of 0.5 M phosphate buffer, pH 6.7. You have available both NaH2PO4 and Na2HPO4.
2) Prepare 750 mL of 0.5 M phosphate buffer, pH 7.8. You have available Na2HPO4, 6 M NaOH and 6 M HCl.
3) Prepare 750 mL of 0.5 M phosphate buffer, pH 7.2. You have available NaH2PO4, 6 M NaOH and 6 M HCl.
4) Prepare 1.5 L of 300 mM Tris buffer, pH 8.2. You have available both Tris base and Tris-HCl.
5) Prepare 1.5 L of 300 mM Tris buffer, pH 7.8. You have available Tris base, 6 M NaOH and 6 M HCl.
6) Prepare 1.5 L of 300 mM Tris buffer, pH 8.8. You have available Tris-HCl, 6 M NaOH and 6 M HCl.
7) Prepare 500 mL of 3.0 M acetate buffer, pH 5.0. You have available both glacial acetic acid and sodium acetate (note: glacial acetic acid is 17.5 M).
8) Prepare 1L of 50 mM phosphate buffer at pH 7.50 using only Na2HPO4 and 1M HCl.
Explanation / Answer
1. The buffer action due to NaH2PO4 and Na2HPO4 is due to inter conversion of the acid H2PO4- to the salt HPO42- .
H2PO4-(aq) + H2O(l) ------> HPO42- (aq) + H3O+ , pKa = 7.21
The required pH = 6.7
Applying Hendersen equation
pH = pKa + log [salt] / [acid] = pKa + log [HPO42-(aq)] / [H2PO4-(aq)]
=> 6.7 = 7.21 + log [HPO42-(aq)] / [H2PO4-(aq)] ----- (1)
Let's represent HPO42-(aq) as B and H2PO4-(aq) as A
=> > 6.7 = 7.21 + log[B] / [A]
=> log[B] / [A] = 6.7 - 7.21 = - 0.51
=> [B] / [A]= antilog (- 0.51) = 10-0.51 = 0.309
=> [B] / [A] = 0.309 --------- (2)
Since volume (V= 750 mL) is same for both A and B, from eqn(2)
[(moles of B) / V] / [(moles of A) / V] = 0.309
=> (moles of B) / (moles of A) = 0.309
=> (moles of B) = 0.309x(moles of A) ---------- (3)
We need to Prepare 750 mL of 0.5 M phosphate buffer.
V = 750 mL = 750 mLx(1L / 1000mL) = 0.750L
Hence moles of A and B in the phosphate buffer = VxM = 0.750Lx0.5mol/L = 0.375 mol
Hence (moles of B) + (moles of A)= 0.375 mol ------(4)
=> 0.309x(moles of A) + (moles of A)= 0.375 mol [from eqn(3)]
=> 1.309x(moles of A)= 0.375 mol
=>(moles of A) = 0.375 / 1.309 mol = 0.286 mol
Hence moles of B = 0.375 - (moles of A) = 0.375 - 0.286 = 0.089 mol
Mass of A (NaH2PO4) = 0.286 mol x (119.96g/mol) = 34.31 g
Mass of B(Na2HPO4) = 0.089 mol x (141.96g/mol) = 12.34 g
Hence we need to add 34.31 g of NaH2PO4 and 12.34 g of Na2HPO4 and dilute them to 750 mL to prepare a buffer of pH 6.7 (answer)
In a similar way we can find answers for other parts.