1) Determine the pH of the solution (to 2 decimal points) after the addition of
ID: 928209 • Letter: 1
Question
1)Determine the pH of the solution (to 2 decimal points) after the addition of 7.31 mL of 0.0123 M barium hydroxide (Ba(OH)2) to 93.9 mL of 0.0578 M sodium dihydrogen phosphate (NaH2PO4).
Assume that the volumes are additive and that the 5% approximation is valid.
2)Determine the pH of a solution prepared by mixing 75.4 mL of 0.103 M HEPPS-H (C9H20N2O4S ) to 832 mL of 0.198 M HEPPS (C9H19N2O4S ).
Assume the volumes of the solution are additive and that the 5% approximation is valid.
Report your answer to 2 decimal places.
3)Determine the pH (to two decimal places) of a solution prepared by adding 48.1 mL of 0.584 M phenylammonium chloride (C6H5NH3Cl) to 281 g of phenylamine (aniline) (C6H5NH2).
Assume the volume of the solution does not change and that the 5% approximation is valid.
Explanation / Answer
1) 2H2PO4- + Ba(OH)2 ---> 2HPO4^2- + Ba2+
moles of Ba(OH)2 = 0.0123 M x 0.00731 L = 9 x 10^-5 mols
moles of NaH2PO4 = 0.0578 M x 0.0939 L = 5.43 x 10^-3 mols
[HPO4^2-] = 1.8 x 10^-4/(0.00731+0.0939) = 1.78 x 10^-3 M
remaining [H2PO4-] = 5.25 x 10^-3/(0.00731+0.0939) = 0.052 M
pH = pKa2 + log([HPO4^2-]/[H2PO4-])
= 7.2 + log(1.78 x 10^-3/5.43 x 10^-3)
= 6.716
2) pH = pKa + log([HEPPS]/[HEPPS-H])
[HEPPS] in solution = 0.198 M x 0.832 L/(0.832 + 0.0754) = 0.181 M
[HEPPS-H] in solution = 0.103 M x 0.0754 L/(0.832 + 0.0754) = 8.56 x 10^-3 M
Feed values,
pH = 8.0 + log(0.181/8.56 x 10^-3)
= 9.32
3) moles of PhNH2 = 281g/93.13 g/mol = 3.02 mols
[PhNH2] = 3.02/0.0481 = 62.786 M
[PhNH3Cl] = 0.584 M
pH = pKa + log([PhNH2]/[PhNH3Cl])
= 4.87 + log(62.786/0.584)
= 6.90
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