Question

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 S^2- in a 0.0590 M hydrosulfuric acid solution, H₂S (aq).

For H₂S, K_a1 = 1.00×10^-7 and K_a2 = 1.00×10^-19

pH =
[HS-] =	M?
[S2-] =	M?

Answer 1

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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