Given the information from Table 3. Complete table 4. Please include a sample ca

Question

Given the information from Table 3. Complete table 4. Please include a sample calculation for 1 trial.

Table 3: Data Table for Titration of an Unknown Weak Acid

Preparation of Acid Solution: Total Volume: 250 mL, Total mass: 4.1081g, 25mL of acid + 25mL of DI water used in titration

0.63mL

Table 4: pKa of unknown acid

Titration results Unknown Trial 1 Unknown Trial 2 Unknown Trial 3 Exact Molarity of NaOH Sol. 0.11854M 0.11854M 0.11854M Initial Volume

0.63mL

0.50 mL 0.18mL Final Volume 27.65mL 27.41mL 27.21mL Equivilance (final-initial) 27.02mL 26.91mL 27.03mL pH measured after adding 2nd aliquot of acid to the titrated solution 6.78pH 6.80pH 6.79pH

Explanation / Answer

Moles of NaOH added = 0.11854 M * 0.02702 L = 3.2*10^-3 moles

moles of acid present = 3.2*10^-3 moles

Molarity of acid = 3.2*10^-3 moles/0.05 L = 0.064 M

pH = 6.78

[H+] = 10^-pH = 10^-6.78 = 1.66*10^-7 M

HA <==> H+ A-

Ka = x^2/(0.064-x)

x = [H+] = 1.66*10^-7 M

Ka = (1.66*10^-7 M)/ (0.064-1.66*10^-7 M) = 4.3 *10^-13

--------------------------------------------------------------

From Trial 2.

Ka = 3.91*10^-13

From Trial 3 :

Ka = 4.1*10^-13

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