The dibasic compound G (pK b1 =4.5, pK b2 =8.5) was titrated with 0.500 M HCl. T

Question

1. The dibasic compound G (pKb1=4.5, pKb2=8.5) was titrated with 0.500 M HCl. The initial solution of G was 0.100 M and had a volume of 100.0 mL. Find the pH after 20.0 mL of HCl has been added. A. 5.5 B. 8.0 C. 8.5 D. 7.5 E. 6.5

1. The dibasic compound G (pKb1=4.5, pKb2=8.5) was titrated with 0.500 M HCl. The initial solution of G was 0.100 M and had a volume of 100.0 mL. Find the pH after 20.0 mL of HCl has been added. A. 5.5 B. 8.0 C. 8.5 D. 7.5 E. 6.5

The dibasic compound G (pKb1=4.5, pKb2=8.5) was titrated with 0.500 M HCl. The initial solution of G was 0.100 M and had a volume of 100.0 mL. Find the pH after 20.0 mL of HCl has been added. A. 5.5 B. 8.0 C. 8.5 D. 7.5 E. 6.5 The dibasic compound G (pKb1=4.5, pKb2=8.5) was titrated with 0.500 M HCl. The initial solution of G was 0.100 M and had a volume of 100.0 mL. Find the pH after 20.0 mL of HCl has been added. 5.5 8.0 8.5 7.5 6.5 A. 5.5 B. 8.0 C. 8.5 D. 7.5 E. 6.5

Explanation / Answer

moles of B^2- = molarity x volume = 0.1 x 0.1 = 0.01 mols

moles of HCl added = 0.5 x 0.02 = 0.01 mols

So we would have 0.01 mols of HB- at this stage

molarity of HB- = 0.01/(0.1 + 0.02) = 0.0833 M

HB- + H2O <==> H2B + OH-

let x amount has reacted,

Kb2 = 3.16 x 10^-9 = x^2/0.0833

x = [OH-] = 1.62 x 10^-5 M

pOH = -log[OH-] = 5.0

pH = 14-5 = 9 (approx 8.5)

So the correct answer wold be,

pH of solution,

C. 8.5

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---