Answer:
7.92 mL
Explanation:
Step 1: Write the balanced neutralization equation
HCH₃CO₂ + NaOH → NaCH₃CO₂ + H₂O
Step 2: Calculate the moles of acetic acid
The molar mass of acetic acid is 60.05 g/mol. The moles corresponding to 0.0617 g are:
[tex]0.0617g \times \frac{1mol}{60.05g} =1.03 \times 10^{-3} mol[/tex]
Step 3: Calculate the moles of NaOH
The molar ratio of HCH₃CO₂ to NaOH is 1:1. The reacting moles of NaOH are 1.03 × 10⁻³ mol.
Step 4: Calculate the volume of NaOH
[tex]1.03 \times 10^{-3} mol \times \frac{1L}{0.1300mol} =7.92 \times 10^{-3} L = 7.92 mL[/tex]
Answer:
We have to add 7.92 mL of NaOH
Explanation:
Step 1: Data given
Mass of acetic acid = 0.0617 grams
Volume of the flask = 250.0 mL = 0.250 L
Molarity of NaOH = 0.1300 M
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate moles acetic acid
Moles acetic acid = mass acetic acid / molar mass acetic acid
Moles acetic acid = 0.0617 grams / 60.05 g/mol
Moles acetic acid = 0.00103 moles
Step 4: Calculate concentration CH3COOH
Concentration CH3COOH = moles CH3COOH / volume
Concentration CH3COOH = 0.00103 moles / 0.250 L
Concentration CH3COOH = 0.00412 M
Step 5: Calculate the volume of NaOH needed
C1*V1 = C2*V2
⇒with C1 = the concentration of CH3COOH = 0.00412 M
⇒with V1 = the volume of CH3COOH = 0.250 L
⇒with C2 = the concentration of NaOH = 0.130 M
⇒with V2 = the volume of NaOH needed = TO BE DETERMINED
0.00412 M * 0.250 L = 0.130 M * V2
V2 = (0.00412 M * 0.250 L ) / 0.130 M
V2 = 0.00792 L = 7.92 mL
We have to add 7.92 mL of NaOH
