You are asked to prepare several different buffers. Given below is a list of ava

Question

You are asked to prepare several different buffers. Given below is a list of available materials. For simplicity, you may neglect volume changes which occur upon mixing any two liquids or dissolving any solids. In each case, your buffer should have a reasonable buffering capacity. The buffer component present in the lower concentration (whether it be the acid or base) should be present at exactly 0.100 M (so, the other component must be at that concentration or higher). Design a recipe meeting these criteria to produce 100.0 mL of each of the buffers required below. (It is sufficient to calculate volumes and masses needed; you need not provide the technician with detailed instructions on weighing procedure, glassware used, etc..) Available Solids: benzoic acid (C_6H_5COOH), sodium chlorite (NaClO_2:), potassium hydrogen phosphate trihydrate (K_2HPO_4 middot 3H_2O) Available Solutions: 0.1973 M HCl, 0, 1309 M NaOH, 2.142 M ammonia Buffers required: pH 4.00 pH 7.00 pH 10.00

Explanation / Answer

An idea about the pKa values of the solids will help us determine which solid to choose.

Benzoic acid: pKa = 4.20

Sodium chlorite: pKa = 10-11

Potassium hydrogen phosphate trihydrate: pKa1 = 2.16, pKa2 = 7.21, pKa3 = 12.32

(B) Since we wish to prepare a pH 7.0 buffer, we must select potassium hydrogen phosphate trihydrate. The ionization of phosphoric acid can be shown as:

H3PO4 -------> H+ + H2PO4- ; pKa1 = 2.16

H2PO4- -------> H+ + HPO42-; pKa2 = 7.21

HPO42- -------> H+ + PO43-; pKa3 = 12.32

We need to work with K2HPO4.3H2O and the buffer system will consist of KH2PO4/K2HPO4. Since the desired pH 7.0 is less than the pKa of the buffering system, we need to add acid.

We use the Henderson-Hasslebach equation as

pH = pKa2 + log [K2HPO4]/[KH2PO4]

As we are decreasing the pH by adding acid, we must have [K2HPO4] < [KH2PO4]. It is given that

[K2HPO4] = 0.100 M

Plug in values and obtain

7.0 = 7.21 + log (0.100)/[KH2PO4]

===> -0.21 = log (0.100)/[KH2PO4]

===> (0.100)/[KH2PO4] = antilog(-0.21) = 0.616

===> [KH2PO4] = (0.100)/(0.616) = 0.162

The concentration of KH2PO4 needed for the buffer is 0.162 M (ans); realize that we do not have KH2PO4 available. Instead, we have 0.1973 M HCl which can react with K2HPO4 to produce KH2PO4.

Moles of K2HPO4 = (volume of solution in L)*(concentration in mol/L) = (100 mL)*(1 L/1000 mL)*(0.100 mol/L) = 0.01 moles.

Moles of KH2PO4 = (100 mL)*(1 L/1000 mL)*(0.162 mol/L) = 0.0162 mol.

Write down the neutralization reaction as

K2HPO4 + HCl --------> KH2PO4 + KCl.

Moles KH2PO4 formed = moles HCl added.

Moles HCl added = 0.0162 mole.

Volume of HCl required = moles of HCl/molarity of HCl = (0.0162 mol)/(0.1973 mol/L) = 0.082108 L = 82.108 mL 82.11 mL (ans).

Molar mass of potassium hydrogen trihydrate = 228.22 g/mol.

Therefore, mass of potassium hydrogen trihydrate required = (0.01 mole)*(228.22 g/mol) = 2.2822 g (ans).

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.