Vinegar Titration Hello guys im struggling with the last part and i need help fi
ID: 1067396 • Letter: V
Question
Vinegar Titration
Hello guys im struggling with the last part and i need help finishing it thanks
Data Table 1-Amount of NaOH added Trial Initial volume (mL) Final volume (mL) Change in volume (mL) Moles of NaOH added (mol) 1 98 1.6 8.2 0.0041 2 0.00405 9.5 1.4 8.1 3 9.2 1.2 8 0.004 Data Table 2-Acetic acid in vinegar Volume of vinegar Volume vinegar Concentration of Trial Moles CH3COOH neutralized (mol) used (mL) used (L) CH3COOH in vinegar (M) 1 0.0041 0.004L. 1.025 4mL 2 0.00405 0.004 1.0125 3 0.004 1 0.004Explanation / Answer
Average concentration of CH3COOH stock solution (M) = (1.025 + 1.0125 + 1.00)/3 = 1.0125
Standard deviation of concentration of CH3COOH stock solution (M) =
{{(1.025 – 1.0125)2 + (1.0125 – 1.0125)2 + (1.00 – 1.0125)2}/3] = [(0.0003125)/3] = 0.000104167 =0.010206 0.0102
The standard deviation of the concentration of CH3COOH stock solution is 0.0102 M (ans).
Typical concentration of acetic acid in vinegar (mass %):
The typical concentration of acetic acid in vinegar must be given in your text; you haven’t supplied that information here.
However, look at the footnote; it says to consider 5% (w/w) as the typical concentration of acetic acid in vinegar. Therefore, the typical concentration is 5% (w/w).
We need to find out the concentration of acetic acid in the stock vinegar solution. The molar concentration of the solution is 1.0125 M = 1.0125 mol/L.
This simply means that 1 L of the solution contains 1.0125 mol CH3COOH.
We know that 1 L = 1000 mL and assume the density of the solution to be 1 g/mL (for simplicity sake; no other information is given as well).
Therefore, mass of 1 L solution = volume of solution*density of solution = (1000 mL)*(1 g/1 mL) = 1000 g.
Therefore,
1000 g solution contains 1.0125 mol CH3COOH.
Molar mass of CH3COOH = (12*2 + 16*2 + 4*1) g/mol = 60 g/mol.
Therefore, 1.0125 mol CH3COOH = (1.0125 mol)*(60 g/1 mol) = 60.75 g CH3COOH
Thus,
1000 g solution contains 60.75 g CH3COOH.
Therefore, 100 g solution contains (60.75 g)*(100 g/1000 g) = 6.075 g CH3COOH.
Since the concentration of acetic acid in vinegar is the mass of CH3COOH in 100 g solution, therefore, the mass percent of the stock solution is 6.075% (ans).
Typical concentration of acetic acid in vinegar (M):
Again this information should have been supplied.
We can back track if we assume the typical concentration to be 5%.
Therefore, 100 g solution contains 5 g CH3COOH.
Thus, 1000 g solution will contain (5 g)*(1000 g/100 g) = 50 g CH3COOH.
Since the density of the solution is 1 g/mL, therefore, 1 L solution contains 50 g CH3COOH.
Moles of CH3COOH = (50 g)*(1 mol/60 g) = 0.83 mol CH3COOH.
Therefore, 1 L of solution contains 0.83 mol CH3COOH.
Typical molar concentration is 0.83 mol/L = 0.83 M (ans).
Percent error:
Typical concentration = 5% (w/w)
Stock solution has 6.075% (w/w).
Percent error = (6.075 – 5)/(5)*100 = 21.5% (ans).
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