PrintCakculatorPeriodic Table n 11 of 12 Mapd Sapling Learning You are asked to

Question

PrintCakculatorPeriodic Table n 11 of 12 Mapd Sapling Learning You are asked to prepare 500. mL of a 0.250 M acetate buffer at pH 5.10 using only pure acetic acid (MW-60.05 g/mol, pK,-4.76),3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer 1. How many grams of acetic acid will you need to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution. Number What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 5.10 ata final volume of 500 mL? (ignore activity coefficients.) Number mL the rinses to the flask. Swirl the flask to mix the solution. Add water to the mark to make the volume 500 mL. Stopper the flask and invert how many times to ensure complete mixing? 3. Pour the beaker contents into a 500 mL volumetric flask. Rinse the beaker several times and add 2 20 Check Answer 0 Nox1 ? Exit

Explanation / Answer

1)

Answer

7.5063g

Explanation

No of moles of acetic acid needed = (0.250mol/1000ml)×500ml= 0.125mol

Mass of acetic acid required = 0.125mol×60.05g/mol = 7.5063g

2)

Answer

28.6ml

Explanation

Henderson-Hasselbalch equation is

pH = pKa + log([A-] /[HA])

rearranging equation

[A-]/[HA] = 10pH - pKa

   [A-] /[HA] = 105.10 - 4.76

     [A-] /[HA] = 2.1878

[A-] = 2.1878[HA]

Buffer concentration is 0.250M

Therefore,

[A-] + [HA-] = 0.250M

2.1878[HA]+ [HA] = 0.250M

3.1878[HA] = 0.250M

[HA] = 0.0784M

[A-] = 0.250M - 0.0784M = 0.1716M

So, concentrations of acetic acid and acetate to be maintained are

[CH3COOH] = 0.0784M

[CH3COO-] = 0.1716M

No of moles to be maintained

No of moles of CH3COOH = (0.0784mol/1000ml) × 500ml = 0.0392

No of moles of CH3COO- = (0.1716mol/1000ml)×500ml =0.0858

NaOH + CH3COOH - - - - - - - > CH3COONa + H2O

this is 1:1molar reaction

So, to produce 0.0858moles of CH3COO-, 0.0858moles of NaOH must be added

Volume of NaOH solution contaning 0.0858moles of NaOH = (1000ml /3.00mol)×0.0858mol = 28.6ml

3) 20

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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