Question

A 10.5 mL sample of vinegar, containing acetic acid, was titrated using 0.460 M NaOH solution. The titration required 19.13 mL of the base. What was the molar concentration of acetic acid in the vinegar?

Answer 1

Answer: 0.25248 M

Step-by-step explanation:

molarity equilibrium is aL * bM = cL * dM

10.5 mL = 0.0105 L

19.13 mL = 0.01913 L

0.0105 L * 0.460M = 0.01913 L * x

x =
`(0.0105 L * 0.460M )/(0.01913 L)` = 0.25248 M

round your answer depending on how many sig figs you need

Answer 2

The concentration of acetic acid is 0.838 M

Step-by-step explanation:

Step 1: Data given

Volume of a 0.460 M NaOH solution = 19.13 mL = 0.01913 L

Volume of the acetic acid = 10.5 mL = 0.0105 L

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of acetic acid

b*Ca*Va = a * Cb*Vb

⇒with B = the coefficient of NaOH = 1

⇒with Ca = the concentration of acetic acid = TO BE DETERMINED

⇒with Va = the volume of acetic acid = 0.0105 L

⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.460 M

⇒ with Vb = the volume of NaOH = 0.01913 L

Ca = Cb*Vb / Va

Ca = (0.460 * 0.01913) /0.0105

Ca =0.838 M

The concentration of acetic acid is 0.838 M