1. A mass of 0.1212 g of an unknown acid is dissolved in a total volume of 25.00
ID: 551849 • Letter: 1
Question
1. A mass of 0.1212 g of an unknown acid is dissolved in a total volume of 25.00 mL. The 25 mL sample is titrated with 0.050 M NaOH. The titration data is shown in the Table .
a.Use Excel to draw the titration curve. (already did this)
b.From the graph, estimate the equivalence point and the half-equivalence point. (estimated the equivalence point but not sure if it's right)
c.Use the data to calculate the Molar Mass of the acid
d.Use the Ka table to identify the acid
Titration Curve 12 10 10 30 40 50 70 80 90 100 Volume Added (mL)Explanation / Answer
b. The equivalence pointat pHexp = 11.99 and the half-equivalence point at pHexp = 8.08
c. Using Molarity equation (Molarity = (Weight of acid / Molar mass of acid)* Volume of solution in liters
Macid * Vacid = MNaOH * VNaOH (1L = 1000mL)
((0.1212g / Molar mass of Acid )*1000/ 25 L) * 25 mL = 0.050M *25 mL
Molar mass of acid = (0.1212 *1000 )/0.050 g/mol = 2424 g/mol
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