1. Calculate the pH of a solution made by mixing 50.0 mL of 0.10 M benzoic acid
ID: 587241 • Letter: 1
Question
1. Calculate the pH of a solution made by mixing 50.0 mL of 0.10 M benzoic acid and 100. mL of 0.15 M potassium benzoate.Calculate the pH of a solution made by mixing 50.0 mL of 0.10 M benzoic acid and 100. mL of 0.15 M potassium benzoate.
2. Calculate the pH of a 0.0100 M solution of sulfurous acid
A) Which of the following combinations will give a buffered solution that has a pH of about 5? Explain clearly the reason for your choice. a) NH3 mixed with NH4Cl (Kb for NH3= 1.8 x10R5 ) or b) C5H5N mixed with C5H5NHCl (Kb for C5H5N = 1.7x10R9 )
(B) What ratio of the concentrations of the conjugate acid/base pair from Part A will be needed to form a buffer solution with a pH of 6.2?
Explanation / Answer
1. Calculate the pH of a solution made by mixing 50.0 mL of 0.10 M benzoic acid and 100. mL of 0.15 M potassium benzoate.
solution :
benzoic acid pKa = 4.20
millimoles of benzoic acid = 50 x 0.10 = 5
millimoles of potassium benzoate = 100 x 0.15 = 15
pH = pKa + log [potassium benzoate / benzoic acid ]
pH = 4.20 + log (15 / 5)
pH = 4.68
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