Question

Consider a buffer solution that contains 0.25 M C₆H₄(CO₂H)(CO₂K) and 0.15 M C₆H₄(CO₂K)₂. pK_a(C₆H₄(CO₂H)CO₂^-)=5.41.

1. Calculate its pH

2. Calculate the change in pH if 0.140 g of solid NaOH is added to 190 mL of this solution.

3. If the acceptable buffer range of the solution is ±0.10 pH units, calculate how many moles of H₃O⁺ can be neutralized by 250 mL of the initial buffer.

Answer 1

1) The pH is calculated using the henderson hasselbach equation:

pH = pKa + log ([Salt] / [Acid]) = 5.41 + log (0.15 / 0.25) = 5.19

2) The moles of salt, acid and NaOH added are calculated:

n Salt = M * V = 0.15 * 0.19 = 0.0285 mol

n Acid = 0.25 * 0.19 = 0.0475 mol

n NaOH = g / MM = 0.14 / 40 = 0.0035 mol

The NaOH reacts with the acid (decreasing it) and forms salt (increasing it), the new pH is calculated:

pH = 5.41 + log (0.0285 + 0.0035 / 0.0475 - 0.0035) = 5.27

3) If H3O + is added, the pH decreases, then the minimum bearable pH is 5.19 - 0.10 = 5.09

The molar ratio of salt and acid for said pH is calculated, using the henderson hasselbach equation cleared:

n Salt / n Acid = 10 ^ (pH - pKa) = 10 ^ (5.09 - 5.41) = 0.48

It has:

1) n Salt - 0.48 * n Acid = 0

2) n Salt + n Acid = 0.40 M * 0.25 L = 0.1 mol

System of equations is applied and you have:

n Salt = 0.03 mol

n Acid = 0.07 mol

The initial moles of salt are:

n Initial salt = 0.15 M * 0.25 L = 0.0375 mol

The moles of H3O + supported are calculated:

n H3O + = n Initial salt - n Final salt = 0.0375 - 0.03 = 0.0075 mol

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