1.)
An aqueous solution contains 0.431 M
ethylamine
(C2H5NH2).
How many mL of 0.368 M hydrobromic
acid would have to be added to 225 mL of
this solution in order to prepare a buffer with a pH of
10.400?
2) A buffer solution contains
0.308 M ammonium
bromide and 0.319 M
ammonia.
If 0.0500 moles of perchloric
acid are added to 225 mL of this buffer,
what is the pH of the resulting solution?
(Assume that the volume does not change upon adding
perchloric acid.)
pH = ?
3) A buffer solution contains 0.476 M
hypochlorous acid and 0.234 M
sodium hypochlorite.
If 0.0211 moles of sodium
hydroxide are added to 125 mL of this
buffer, what is the pH of the resulting solution?
(Assume that the volume does not change upon adding sodium
hydroxide).
pH = ?
4.Calculate the pH and the equilibrium concentrations of
HS- and S2-
in a 0.0590 M hydrosulfuric acid
solution, H2S (aq).
For H2S, Ka1 =
1.00×10-7 and Ka2 =
1.00×10-19
pH = | |
[HS-] = | M? |
[S2-] = | M? |
1) The molar ratio between salt and acid is calculated, using the Henry Hasselbach equation cleared:
n Salt / n Acid = 10 ^ (pH - pKa) = 10 ^ (10.4 - 10.63) = 0.59
It has:
1) n Salt - 0.59 * n Acid = 0
2) n Salt + n Acid = M * V = 0.431 * 0.225 = 0.1 mol
System of equations between 1 and 2 is applied and you have:
n Salt = 0.037 mol
n Acid = 0.063 mol
The HBr reacts with the salt (decreasing it) and acidic form (increasing it), then the moles of acid in the buffer, are the moles of the HBr added, the mL of HBr is calculated:
mL HBr = n * 1000 / M = 0.063 * 1000 / 0.368 M = 171.20 mL
2) The moles of the buffer components are calculated:
n Salt = 0.319 * 0.225 = 0.072 mol
n Acid = 0.308 * 0.225 = 0.069 mol
The HClO4 reacts with the salt (decreasing it) and forms acid (increasing it), the pH is calculated:
pH = pKa + log (n Salt / n Acid) = 9.25 + log (0.072 - 0.05 / 0.069 + 0.05) = 8.52
3) Same previous procedure:
n Salt = 0.234 * 0.125 = 0.029 mol
n Acid = 0.476 * 0.125 = 0.060 mol
The hydroxide reacts with the acid (decreasing it) and forms salt (increasing it):
pH = 7.53 + log (0.029 + 0.0211 / 0.06 - 0.0211) = 7.64
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