Submit for Grading Current score: O/3 pts (0%) Tutored Practice Problem 18.2.7 c

Question

Submit for Grading Current score: O/3 pts (0%) Tutored Practice Problem 18.2.7 coNDA Prepare a buffer by acid-base reactions Close Problem Consider how to prepare a buffer solution with pH-7.13 (using one of the weak acid conjugate base systems shown here) by combining 1.00 L of a 0.425-M solution of weak acid with 0.379 M potassium hydroxide Weak Acid Conjugate Base K HNO2 HCIO HCN 4.5x 10-4 3.5 x 10-8 4.0 x 10-10 pK, 3.35 7.46 9.40 NO2 CIO. How many L of the potassium hydroxide solution would have to be aded to the acid solution of your choicer? Show Approach Alpha Plots As shown in the preceding example problems, the pH of a buffer solution is controlled by the relative amounts of weak acid and conjugate base present and by the weak acid p.k,. That is, buffer pH is of solution volume If a buffer solution is diluted, the pHi does not change Regardless of the

Explanation / Answer

HClO/ClO- is suitable

Henderson-Hasselbalch equation is

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

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

[A-] / [HA] = 107.13 - 7.46

[A-] / [HA] = 0.4677

[A-] = 0.4677×[HA]

Total buffer concentration = 0.425M

so,

[A-] + [HA] = 0.425M

0.4677[HA]+ [HA] = 0.425M

1.4677[HA]= 0.425M

[HA] = 0.2896M

[A-] = 0.425M - 0.2896M = 0.1354M

Therefore,

the required concentrations

[HClO] = 0.2896M

[ClO-]= 0.1354M

Now, consider the reaction between HOCl and KOH

HClO + KOH - - - - - > KClO + H2O

stoichiometrically, 1mole of KOH react with 1mole of HClO to give 1 mole of ClO-

No of moles of ClO- required = 0.1354

No of moles of KOH required = 0.1354

Volume of KOH required = (1L/0.379mol)×0.1354 = 0.3573L

Therefore,

0.3573L of 0.379M KOH solution must be added to 1L of 0.425M HClO solution.

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