Question

. You transfer 25.00 mL of your Kroger brand vinegar solution via volumetric pipet to a 250.00 mL volumetric flask and dilute to the final volume using distilled water, after which you mix the solution well. Next, you take a 25.00 mL aliquot of this diluted commercial vinegar solution and transfer it to a 150 mL Erlenmeyer flask. Titration of this sample to the phenolphthalein endpoint with sodium hydroxide required 15.81 mL of the 0.1002 M NaOH titrant. Based on this data, what is the molar concentration of the acetic acid in the original Kroger brand commercial vinegar solution ?

Answer 1

Answer:

0.0634 M.

Explanation:

Equation of the neutralisation reaction:

CH3COOH + NaOH--> CH3COONa + H2O

Number of moles = molar concentration * volume

= 0.1002 * 0.01581

= 0.00158 mol.

By stoichiometry,

1 mole of acetic acid reacts with 1 mole of NaOH

Number of moles of acetic acid = 0.00158 mol.

Concentration in 2nd dilution = moles/ volume

= 0.00158/0.25

= 0.00634 M

At 25 ml,

Concentration =

C1 * V1 = C2 * V2

= (0.00634 * 0.25)/0.025

= 0.0634 M.

Answer 2

The molar concentration will be "0.0634 M".

Given:

• Molar concentration of NaOH = 0.1002 M
• Volume = 15.81 mL or, 0.01581 L

Neutralization reaction's equation:

• $CH_3 COOH+ NaOH \rightarrow CH_3COONa+H_2O$

Now,

The number of moles will be:

= $Molar \ concentration\times Volume$

= $0.1002\times 0.01581$

= $0.00158 \ mol$

and,

The concentration of second dilution will be:

= $\frac{Moles}{Volume}$

= $\frac{0.00158}{0.25}$

= $0.00634 \ M$

hence,

The concentration at 25 mL will be:

→ $C_1\times V_1 = C_2\times V_2$

or,

→         $C_2 = \frac{C_1\times V_1}{V_2}$

                 $= \frac{0.00634\times 0.25}{0.025}$

                 $= 0.0634 \ M$

Thus the above response is right.  

Learn more about molarity here:

https://brainly.com/question/13599639