Question 6 (10 pts) Buffers. Find the pH for the following buffers. Give a balan
ID: 580420 • Letter: Q
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Question 6 (10 pts) Buffers. Find the pH for the following buffers. Give a balanced reaction equation in all cases. Use the Henderson-Hasselbalch equation to solve these 3 problems!! a) A buffer consisting of 0.12 M sodium carbonate and 0.82 M sodium bicarbonate (Ka = 4.7 x 10-11) b) What is the pH of 500 mL of solution containing 0.125 mol/L of propanoic acid (HPr) and 0.375 mol/L of sodium propanoate (NaPr)? (Ka for propanoic acid is 5.52 x 10.) A frequently used buffer in biological systems, including IV fluids, is phosphate- buffered saline ("PBS"), which consists of 1.44 g of Na2HPO4.2H20, 0.24 g of KH2PO, and 8.00 g of NaCl per liter of solution. What is its pH? (It is a very good idea to write out the reaction for this equilibrium first! See Table 15.5, p.683, for the Ka values of the phosphates.) c)Explanation / Answer
a) Na2CO3 ----> 2Na+ + CO32-
NaHCO3 ----> Na+ + HCO3-
According to Handerson Hasselbach equation,
pH = pKa + log [salt]/[acid]
Ka = 4.7*10-11
pKa = -log Ka = -log(4.7*10-11) = 10.33
pH = pKa + log [CO32-]/[HCO3-]
= 10.33 + log (0.12/0.82)
= 10.33-0.83 = 9.50
pH = 9.50
b) NaPr ----> Na+ + Pr-
According to Handerson Hasselbach equation,
pH = pKa + log [salt]/[acid]
Ka = 5.52*10-5
pKa = -log Ka = -log(5.52*10-5) = 4.26
[HPr] = 0.125 M
[Pr-] = 0.375 M
pH = pKa + log [Pr-]/[HPr]
= 4.26 + log (0.375/0.125)
= 4.74
pH = 4.74
c) Na2HPO4 -----> 2Na+ + HPO42-
KH2PO4 ----> K+ + H2PO4-
NaCl -----> Na+ + Cl-
NaCl will not contribute to the pH of the solution as it is a salt.
Mass of Na2HPO4.2H2O = 1.44 g
Molar mass of Na2HPO4.2H2O = 178 g/mol
moles of Na2HPO4.2H2O = mass/molar mass = 1.44/178 = 0.0081 mol
Mass of KH2PO4 = 0.24 g
Molar mass of KH2PO4 = 136 g/mol
moles of KH2PO4 = mass/molar mass = 0.24/136 = 0.0017 mol
According to Handerson Hasselbach equation,
pH = pKa + log [salt]/[acid]
Ka = 6.3*10-8
pKa = -log Ka = -log(6.3*10-8) = 7.20
pH = pKa + log [HPO42-]/[H2PO4-]
= 7.20 + log (0.0081/0.0017)
= 7.88
pH = 7.88
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