Answer:
28.6 mL
Explanation:
Step 1: Write the balanced neutralization reaction between acetic acid and sodium hydroxide
CH₃COOH(aq) + NaOH(aq) = CH₃COONa(aq) + H₂O(l)
Step 2: Calculate the moles of acetic acid
The molar mass of acetic acid is 60.05 g/mol. The moles corresponding to 0.120 g are:
[tex]0.120g \times \frac{1 mol}{60.05g} =2.00 \times 10^{-3} mol[/tex]
Step 3: Calculate the moles of sodium hydroxide
The molar ratio of CH₃COOH to NaOH is 1:1. The reacting moles of NaOH are 2.00 × 10⁻³ mol.
Step 4: Calculate the volume of the 0.0700 M NaOH solution
[tex]2.00 \times 10^{-3} mol \times \frac{1L}{0.0700mol} =0.0286 L = 28.6 mL[/tex]
Answer:
We have to add 286 mL of NaOH
Explanation:
Step 1: Data given
Mass of acetic acid (CH3COOH)= 0.120 grams
Volume of acetic acid = 250 mL = 0.250 L
Molarity of NaOH = 0.0700 M
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate moles acetic acid
Moles acetic acid = mass / molar mass
Moles acetic acid = 0.120 grams / 60.05 g/mol
Moles acetic acid = 0.00200 moles
Step 4: Calculate molarity acetic acid
Molarity acetic acid = moles / volume
Molarity acetic acid = 0.00200 moles /0.250 L
Molarity acetic acid = 0.008 M
Step 5: Calculate the volume of solution the student will need to add to reach the equivalence point
C1*V1 = C2*V2
⇒with C1 = the molarity of acetic acid = 0.008 M
⇒with V1 = the volume of acetic acid = 0.250 L
⇒with C2 = the molarity of NaOH = 0.0700 M
⇒with V2 = the volume of NaOH neede = TO BE DETERMINED
0.008 M * 0.250 L = 0.0700 M * V2
V2 = (0.008M * 0.250 L) / 0.0700 M
V2 = 0.286 L = 286 mL
We have to add 286 mL of NaOH
