Table 1 Composition of Solution pH 1= 2.96, 2= 3.13, 3=4.08 A) Write each observ
ID: 973307 • Letter: T
Question
Table 1 Composition of Solution
pH 1= 2.96, 2= 3.13, 3=4.08
A) Write each observed pH
B) Determine the [H3O] concentration using the given pH
C) Determine the degree of ionization (or percent ionization)
Solution Composition 1 0.10 M CH3COOH 2 5 mL 0.10 M CH3COOH 3 1 mL 0.10 M CH3COOH + 99 mL H20 4 5 mL 0.10 M CH3COOH + 5 mL 0.10 M HCl 5 0.10 M H3PO4 6 0.10 M NH3 7 0.10 M NH4NO3 8 25 mL 0.10 M NH3 + 25 mL 0.10 M NH4NO3 9 10 mL Soln 8 + 1 mL 0.10 M HCl 10 10 mL Soln 8 + 1 mL 0.10 NaOH 11 10 mL Soln 8 + 6 mL 0.10 M HClExplanation / Answer
(A) Observed pH for solution 1 = 2.96
solution 2 = 3.13
solution 3 = 4.08
(B) Solution 1 [H3O+] = 10 -2.96 = 0.0011M
Solution 2 [H3O+] = 10 -3.13 = 0.00074M
SOlution 3 [H3O+] = 10 -4.08 = 0.0000832M
Part (C)
Solution 1 degree of ionization = (Ka / C )1/2 = (1.8 x 10 -5 / 0.1)1/2 = 0.01342 = 1.3 %
Solution 2 degree of ionization =(Ka / C )1/2 = (1.8 x 10 -5 / (0.1x5/1000))1/2 = 0.190= 19 %
Solution 3 degree of ionization =(Ka / C )1/2 = (1.8 x 10 -5 / (0.1x1/100))1/2 = 0.134 = 13.4 %
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