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?

0.25248 M

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 = = 0.25248 M

round your answer depending on how many sig figs you need

The concentration of acetic acid is 0.838 M

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

b..

is the correct one

explanation:

covalent bonds form as valence electrons are shared between two atoms. lewis structures and structural formulas are common ways of showing the covalent bonding in organic molecules. formal charge describes the changes in the number of valence electrons as an atom becomes bonded into a molecule.

just did the assignment and the answer was #1 +2 n