2.A buffer solution contains 0.295 M C6H5NH3Br and 0.339 M C6H5NH2 (aniline). De
ID: 1010113 • Letter: 2
Question
2.A buffer solution contains 0.295 M C6H5NH3Br and 0.339 M C6H5NH2 (aniline). Determine the pH change when 0.076 mol HBris added to 1.00 L of the buffer. pH after addition pH before addition = pH change =
3.An aqueous solution contains 0.425 M methylamine (CH3NH2). How many mL of 0.365 M hydroiodic acid would have to be added to 250 mL of this solution in order to prepare a buffer with a pH of 10.400?
4.An aqueous solution contains 0.323 M hydrofluoric acid. How many mL of 0.242 M sodium hydroxide would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 3.050?
5.How many grams of solid potassium fluoride should be added to 0.500 L of a 5.59×10-2 M hydrofluoric acid solution to prepare a buffer with a pH of 3.699 ? grams potassium fluoride =
6.How many grams of solid ammonium chloride should be added to 1.00 L of a 0.256 M ammonia solution to prepare a buffer with a pH of 10.180 ? grams ammonium chloride
Explanation / Answer
2. For buffer,
pH = pKa + log(base/acid)
Initial pH = 3.983 + log(0.339/0.295) = 4.043
After 0.076 mol HBr is added
New [C6H5NH3Br] = (0.295 M x 1 L + 0.076) mol/1 L = 0.371 M
New [C6H5NH2] = (0.339 M x 1 L - 0.076) mol/1 L = 0.263 M
New pH = 3.983 + log(0.263/0.371) = 3.833
pH change = 3.833 - 4.043 = -0.21
3. let x amount of HI is added
pH = pKa + log(base/acid)
10.40 = 10.62 + log((0.425 M x 0.25 L - x)/x)
0.106 - x = 0.60x
x = 0.066 mol
Volume of HI to be added = 0.066 mol/0.365 M = 0.1808 L = 180.82 ml
4. let x amount of NaOH is added
pH = pKa + log(base/acid)
3.050 = 3.17 + log(x/0.323 M x 0.125 L - x)
0.03 - 0.76x = x
x = 0.017 mol
Volume of 0.242 M NaOH to be added = 0.017 mol/0.242 M = 0.070 L = 70.25 ml
5. let x amount of KF is added
pH = pKa + log(base/acid)
3.699 = 3.17 + log(x/0.0559 M x 0.5 L - x)
0.0945 - 3.38x = x
x = 0.0216 mol
grams of NaF to be added = 0.0216 mol x 41.99 g/mol = 0.907 g
6. let x amount of NH4Cl is added
pH = pKa + log(base/acid)
10.18 = 9.26 + log(0.256 M x 1 L - x)/x)
0.256 - x = 8.32x
x = 0.027 mol
grams of NH4Cl to be added = 0.027 mol x 53.491 g/mol = 1.47 g
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