Use the References to access important values if needed for this question Design

Question

Use the References to access important values if needed for this question Design a buffer that has a pH of 7.74 using one of the weak acid/conjugate base systems shown below. Weak Acid Conjugate BaseK pKa 6.4 x 10 4.19 6.2x i0" | 7.21 4.8 x 10 10.32 2- HPO,2 CO,2 How many grams of the sodium salt of the weak acid must be combined with how many grams of the sodium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base? grams sodium salt of weak acid | grams sodium salt of conjugate base- Submit Answer Retry Entire Group 8 more group attempts remaining

Explanation / Answer

Ans. Part 1: Choice of weak acid- conjugate base pair: A buffer has maximum buffering capacity when pH of the buffer is closest to the pKa of weak acid.

Among the given options, the pKa of weak acid H2PO4- (7.21) is closest to the given buffer pH of 7.74. So, Phosphate buffer is the best choice to prepare this buffer.

Part 2: Step 1: Preparation of Phosphate Buffer: One mole of each salt dissociates in water to yield 1 mol of their respective phosphate ions as follow-

Na2HPO4 ------> HPO42-(aq) + 2Na+(aq)             - monoprotic, Conjugate Base

NaH2PO4 ------> H2PO4-(aq) + Na+(aq)               - diprotic form, weak acid

The diprotic phosphate H2PO­2- acts as weak acid (pKa = 7.21). It can donate a proton to become HPO42- as follow-

            H2PO4-(aq) <---------> HPO42-(aq) + H+

Now, using Henderson- Hasselbalch equation

            pH = pKa + log ([A-] / [AH])                    - equation 1

                        where, [A-] = [HPO42-] = [Na2HPO4]                      ;

[AH] = [H2PO4-] = [NaH2PO4] = 1.0 M

Now,

            Putting the values in equation 1-

            7.74 = 7.21 + log ([Na2HPO4] / [NaH2PO4])

            Or, 7.74 – 7.21 = log ([Na2HPO4] / 1.0]

            Or, antilog (0.53) = [Na2HPO4]

            Or, [Na2HPO4] = 100.53 = 3.39

Therefore, required [Na2HPO4] = 3.39 M

# Step 2: Now,

Mass of [Na2HPO4] required = (Molarity x Vol. of solution in L) x Molar mass

                                                            = (1.0 M x 1.00 L) x (141.958838 g/mol)

                                                            = 141.959 g

Mass of [NaH2PO4] required = (Molarity x vol. of solution in L) x Molar mass

                                                            = (3.39 M x 1.0 L) x (119.97701 g/mol)

                                                            = 406.722 g

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

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