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4. The stockroom contains 1.0 M NaAc (Sodium Acetate), 1.0 M HAc (acetic Acid), distilled water and strong acids and bases. You need to create a buffer solution that has a pH of 4.75 such that when 1.00 mL of 10.0 M HCl is added to 100.00 mL of your buffer, the resulting pH is 3.75 (+/- 0.1). What concentrations of HAc and NaAc do you need to create the buffer solution? Show your calculations and also write out steps on how you "virtually" performed the titration.

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Answer:

Concentrations of HAc and NaAc you need are 0.122M

Explanation:

pKa of acetic acid is 4.75, that means when amount of sodium acetate and acetic acid is the same, pH will be 4.75

Thus, you know [NaAc]i = [HAc]i

Now, using H-H equation, when pH = 3.75:

3.75 = 4.75 + log [NaAc] / [HAc]

0.1 = [NaAc] / [HAc]

10 [NaAc] = [HAc]

Thus, after the reaction  [HAc] must be ten times,  [NaAc].

Based in the reaction of NaAc with HCl

NaAc + HCl → HAc + NaCl

Moles of HCl added are:

1mL = 0.001L * (10mol /L) = 0.01 moles HCl.

That means moles of both compounds after the reaction are:

[NaAc] = [NaAc]i - 0.01 mol

[HAc] = [HAc]i + 0.01

Replacing these equations with the information you know:

[NaAc] = [NaAc]i - 0.01 mol

10[NaAc] = [NaAc]i + 0.01

Subtracting both equations:

9[NaAc] = 0.02mol

[NaAc] = 0.0022 moles.

Replacing in [NaAc] = [NaAc]i - 0.01 mol

0.0022mol = [NaAc]i - 0.01 mol

0.0122mol = [NaAc]i = [HAc]i

These moles in 100.00mL = 0.1000L:

[NaAc]i = [HAc]i = 0.0122mol / 0.100L =

0.122M

Thus, concentrations of HAc and NaAc you need are 0.122M

To create this buffer, you need to pipette 12.2mL of both 1.0M NaAc and 1.0M HAc and dilute this mixture to 100.0mL

mickymike92

Based on the Henderson-Hasselbach equation, the concentrations of HAc and NaAc required is 0.122M

In order to prepare this buffer, pipette 12.2 mL each of the stock 1.0M NaAc solution and 1.0M HAc solution, and dilute this mixture to 100.0mL.

What is a buffer solution?

A buffer solution is a solution which resists changes to its pH when a small volume of acid or base is added to it.

A buffer solution is usually prepared from a weak acid and its conjugate base or a weak base and its conjugate acid.

From the data provided:

pKa of acetic acid is 4.75

Thus, at pH 4.75, the initial concentrations of NaAc and HAc are equal.

  • [NaAc]i = [HAc]i

From the Henderson-Hasselbach equation, when pH = 3.75

3.75 = 4.75 + log [NaAc]/[HAc]

0.1 = [NaAc] / [HAc]

10[NaAc] = [HAc]

Therefore, after the reaction, [HAc] will be ten times [NaAc].

HCl is added to the buffer to bring the pH to 3.75

Equation of the reaction of NaAc with HCl is given as:

  • NaAc + HCl → HAc + NaCl

  • Moles of HCl added = molarity × volume in L

moles of HCl added = 0.001L * (10mol /L) = 0.01 moles HCl.

Thus, after the addition of HCl, concentration of NaAc and HAc will be:

[NaAc] = [NaAc]i - 0.01 mol

[HAc] = [HAc]i + 0.01

Recall; 10[NaAc] = [HAc]

Substituting [HAc] = 10[NaAc

[NaAc] = [NaAc]i - 0.01 mol ----(1)

10[NaAc] = [NaAc]i + 0.01 -----(2)

Subtracting (1) from (2) equations:

9[NaAc] = 0.02mol

[NaAc] = 0.0022 moles.

Substituting in [NaAc] = [NaAc]i - 0.01 mol

0.0022mol = [NaAc]i - 0.01 mol

[NaAc]i = 0.0122mol

Also, [HAc]i = 0.0122mol

What is the Molarity of the required solutions?

The molarity is determined using the formula below:

  • Molarity of the solution = moles/volume

volumeof buffer = 100 mL = 0.100 L

[NaAc]i = [HAc]i = 0.0122mol / 0.100L

[NaAc]i = [HAc]i = 0.122M

Therefore, concentrations of HAc and NaAc required is 0.122M

In order to prepare this buffer, pipette 12.2 mL each of the stock 1.0M NaAc solution and 1.0M HAc solution, and dilute this mixture to 100.0mL.

Learn more about buffer solutions at: https://brainly.com/question/22390063

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