We did a lab comparing titrated HCl and NaOH with water vs a buffer (Acetic Acid

Question

We did a lab comparing titrated HCl and NaOH with water vs a buffer (Acetic Acid). Based on my calculations, how can I explain the drop in the concentrations of the buffer for both HCl and NaOH? These questions are underneath both calculations.
Thank you! Goal #3: Prepare an acetate buffer. You will make a buffer with the following properties: total volume 100.0 mL; acetic acid concentration 0.100 M; and acetate concentration 0.100 M. Note: you are supposed to make a single solution with these concentrations, not two separate solutions. You will be given the acetic acid in a form of a 1.00 M solution. So you will have to determine the volume of this stock solution to pipet into a 100 ml volumetric flask to achieve the desired concentration. You will be given the acetate in the form of solid sodium acetate trihydrate (CH3COONa 3H20, Mw 136.1 g/mol). Determine the mass of the solid needed to make the appropriate acetate concentration i J Make sure you mix throughout the dilution process. 5-4.7

Explanation / Answer

What I get from the question is that you are first supposed to measure out (either using pipette or by direct weighing) stock acetic acid and sodium acetate.

You have a 1.0 M stock acetic acid solution and you wish to prepare 100.0 mL of 0.10 M acetic acid solution.

Use the dilution equation:

M1*V1 = M2*V2 where M1 = 1.0 M, M2 = 0.1 M, V2 = 100.0 mL; we need to determine V1.

===> (1.0 M)*V1 = (0.1 M)*(100.0 mL)

===> V1 = (0.1*100.0)/(1.0) mL = 10.0 mL.

Take 10.0 mL of 1.0 M stock acetic acid solution and dilute with water till the 100.0 mL mark.

We will prepare 0.1 M sodium acetate solution by direct weighing.

Moles of sodium acetate in 100.0 mL of 0.1 M buffer solution = (100.0 mL)*(1 L/1000 mL)*(0.1 mol/L) = 0.01 mole.

Mass of sodium acetate taken = (moles of sodium acetate)*(molar mass of sodium acetate) = (0.01 mole)*(136.1 g/mol) = 1.361 g (ans).

Weigh out 1.361 g sodium acetate and put in the prepared 100.0 mL buffer solution.

Determine the pH of the prepared buffer solution. Note that the pKa of acetic acid is 4.76. Use the Henderson-Hasslebach equation as

pH = pKa + log [sodium acetate]/[acetic acid] = 4.76 + log (0.1)/(0.1) = 4.76 (ans).

Next you determine the molarity of acetic acid and sodium acetate by titrating with NaOH and HCl.

Consider the titration of sodium acetate (NaA) with HCl as below:

NaA + HCl -----> HA + NaCl

Moles of HCl added = moles of NaA titrated.

Moles of HCl added = (19.0 mL)*(0.135 mol/L) = 2.565 mmole

Moles of NaA titrated = 2.565 mmole.

Volume of NaA taken = 25.0 mL; therefore, concentration of NaA = (2.565 mmole)/(25.0 mL) = 0.1026 M (ans).

The obtained molar concentration is slightly higher than the desired concentration of 0.1 M (ans).

Consider the titration of acetic acid (HA) with NaOH as below:

HA + NaOH -----> NaA + H2O

Moles of NaOH added = moles of HA titrated.

Moles of NaOH added = (19.50 mL)*(0.125 mol/L) = 2.4375 mmole

Moles of HA titrated = 2.4375 mmole.

Volume of HA taken = 25.0 mL; therefore, concentration of HA = (2.4375 mmole)/(25.0 mL) = 0.0975 M (ans).

The obtained molar concentration is slightly lower than the desired concentration of 0.1 M (ans).

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