Pre-Lab Questions 1. Answer the following regarding the titration of 25.00 mL of

Question

Pre-Lab Questions 1. Answer the following regarding the titration of 25.00 mL of a 0.070 M weak acid solution with 0.10 M NaOH. a. How many mL of 0.10 M NaOH are required to reach the first equivalence point? b. How many total mL of 0.10 M NaOH are required to reach the second equivalence point? What is the relationship between the volume of NaOH needed to reach the first and second equivalence points? c. 2. A0.1310 g sample of an unknown diprotic acid is diluted to 100.00 mL and titrated by using 0.1910 M NaOH. If 14.20 mL of the NaOH solution is required to reach the second equivalence point, what is the molar mass of the acid? Use the titration data in the graph below to calculate K,: and Kaz. 3. 12 10 0 15 20 25 10 Volume NaOH (mL)

Explanation / Answer

Q1.

a)

From the graph shown above,

V = 10 mL is required for 1st point

b)

total V, is the 2nd point, which is V = 20 mL ( 2nd drastic change in pH)

c)

ratio between V1eq and V2eq

V2eq/V1eq = 20/10 = 2x

therefore, ratio is 2:1

Q2.

m = 0.1310 g of diprotic acid...

mmol of NaOH = MV = 14.20*0.1910 = 2.7122 mmol of base

mmol of acid = mmol of NaOH = 2.7122

MW = mass/mmol = (0.1310)/(2.7122*10^-3) = 48.30 g/mol

Q3.

Ka1 --> form half equivalence point V = 1/2* 10 = 5 mL

pH at V = 5 --> 2

therefore

pH = pKa + log(a-/HA); at half point A- = HA = 1

pH = pKa1 = 2

KA1 = 10^-pKa1 = 10^-2

for Ka2:

V half --> 15 mL

pH at 15 mL, 6

pH = pKa2 = 6

Ka2 = 10^-pKa2 = 10^-6

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