AP Vinegar Analysis Desk No. Date _labSec. Name 1. Assuming the density of a 5%

Question

AP Vinegar Analysis Desk No. Date _labSec. Name 1. Assuming the density of a 5% acetic acid by mass solution is 1.0 g/mL, determine the volume of the acetic acid solu- don necessary to neutralize 25.0 mL of 0.10 M NaOH. Also record this calculation on your Report Sheet 2. a. A chemist often uses a white card with a black mark to aid in reading the meniscus of a clear liquid. How does this technique make the reading more accurate? Explain. b. A chemist should wait 10-15 seconds after dispensing a volume of titrant before a reading is made. Explain why the wait is good laboratory technique. 164 : c. The color change at the endpoint should persist for 30 seconds. Explain why the time lapse is a good titration technique. 3· Lemon juice has a pH of about 2.5. Assuming that the acidity of lemon juice is due then the citric acid concentration calculates to 0.5% by mass. solely to citric acid, that citric acid is a monoprotic acid, and that the density of lemon Juice is 1.0 g/mL, Estimate the volume of 0.0100 M NaOH required to neutralize a 3.71-g sample of lemon juice. The molar mass of citric acid is 190.12 g/mol. Experiment 10 145

Explanation / Answer

Vol of NaOH = 25.00 ml = 0.025L

Molarity of NaOH = 0.10 M

Moles NaOH = Molarity/Volume(L) = 0.10/0.025 = 4 moles NaOH

NaOH + CH3COOH ------------> CH3COONa + H2O

From the equation, to neutralize 4 moles of NaOH, we would need 4 Moles of CH3COOH.

5% mass solution of CH3COOH is same as 5gms of CH3COOH in 100 ml of H2O (as density is 1.00 g/ml)

Converting % mass into moles:

5g CH3COOH X [(1mol CH3COOH)/(60.00 g CH3COOH)] = 0.0833 moles of CH3COOH

So for evry 100ml of solution, there are 0.0833 moles of CH3COOH. to get 4 moles of CH3COOH, you need (4/0.0833 = 48) 4800 ml or 4.8 L of 5% mass solution of CH3COOH

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---