Lab%20AssignmentE2021ptt /Mateo Marshal/Downloads/Acds%20Bases%205 al sand Buffr

Question

Lab%20AssignmentE2021ptt /Mateo Marshal/Downloads/Acds%20Bases%205 al sand Buffrs Pre-Lab Assignment docx More reasonable, eh? CH.cooH/CH CDO buffer of ph 51. of CtsCOOH-1.80x 13. Sppase you are in the lab and need to prepare a CHCOOOHOO 103 Outline a plan to do this under the following restrictions a. You must prepare exadtly 100.0 ml of the buffer solution. b. You wil use 1.00 Macetic acd as the source of acetic add AND you will make the buffer solution 0.01 Min acetic acid. (You will have to determine the volume of the 1 Macetic acid needed.) c You wil need to calculate the mass of sodum acetate to use.

Explanation / Answer

Use the Henderson-Hasselbalch equation to calculelate ratio of [CH3COO-] /[CH3COOH]

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

rearranging equation

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

Ka of Acetic acid = 1.80×10-5

pKa = - logKa

pKa = - log(1.8×10-5) = 4.75

  substituting the values of pH and pKa

[A-] /[HA] = 105.10 - 4.75

   [A-] / [HA] = 2.2387

so, the ratio of [CH3COO-] /[CH3COOH] = 2.2387

Concentration of CH3COOH= 0.01M

Therefore,

[CH3COO-] = 2.2387× 0.01M = 0.02239M

Now, Calculate volume of 1M CH3COOH needed and weight of CH3COONa needed

C1×V1 = C2× V2

V1 = C2×V2/C1

V1 = 0.01M×100ml/1M

V1 = 1ml

Volume of 1M CH3COOH needed = 1ml

No of moles of CH3COONa needed = (0.02239mol/1000ml)×100ml = 0.002239mol

Molar mass of CH3COONa = 82.03g/mol

Mass of CH3COONa required = 82.03g/mol × 0.002239mol = 0.1837g

So,

pipette out 1ml of 1M acetic acid and transfer it into 100ml volumetric flask, accurately weigh 0.1837g of sodium acetate and transfer it into the volumetric flask, dissolve the content by deionised water and make up to the mark.

  

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