A 10.0 mL aliquot of 0.1000 M base B (pKb1= 4.00, pKb2= 9.00) was titrated with...

A 10.0 mL aliquot of 0.1000 M base B (pKb1= 4.00, pKb2= 9.00) was titrated with 0.1000 M HCl. At what volume is the second equivalence point?

Find the pH at the volume of the acid added (VHCl) equal 10.0 mL

Find the pH at the volume of the acid added (VHCl) equal 20.0 mL.

Please Explain!

Expert Solution

queen_honey_blossom answered 1 year ago

The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl....

The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added and make a graph of pH versus Va; Va=0, 1, 5, 9, 10, 11, 15, 19, 20, and 22mL.

The dibasic compound B (pKb1 5.00, pKb2 9.00) was titrated with 0.5 M HCl. The initial...

The dibasic compound B (pKb1 5.00, pKb2 9.00) was titrated with 0.5 M HCl. The initial solution of B was 0.0500 M and had a volume of 50.0 mL. Find the pH at the following volumes of acid added Va = 0, 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL.

The dibasic compound B (pKb1 = 4.03, pKb2 = 8.06) was titrated with 1.12 M HCl....

The dibasic compound B (pKb1 = 4.03, pKb2 = 8.06) was titrated with 1.12 M HCl. The initial solution of B was 0.112 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added: Va = 0.00, 2.00, 5.00, 9.00, 10.00, 11.00, 15.00, 19.00,20.00, and 22.00 mL. (Assume Kw = 1.01 ✕ 10−14.)

The dibasic compound B (pKb1 = 4.03, pKb2 = 8.06) was titrated with 1.18 M HCl....

The dibasic compound B (pKb1 = 4.03, pKb2 = 8.06) was titrated with 1.18 M HCl. The initial solution of B was 0.118 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added: Va = 0.00, 2.00, 5.00, 9.00, 10.00, 11.00, 15.00, 19.00, 20.00, and 22.00 mL. (Assume Kw = 1.01 ✕ 10−14.)

A 110.0-mL aliquot of 0.120 M weak base B (pKb = 4.55) was titrated with 1.20...

A 110.0-mL aliquot of 0.120 M weak base B (pKb = 4.55) was titrated with 1.20 M HClO4. Find the pH at the following volumes of acid added: Va = 0.00, 1.30, 5.50, 10.00, 10.90, 11.00, 11.10, and 14.00 mL. (Assume Kw = 1.01 ✕ 10−14.)

The pKb values for the dibasic base B are pKb1 = 2.1 and pKb2 = 7.5....

The pKb values for the dibasic base B are pKb1 = 2.1 and pKb2 = 7.5. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.95 M B(aq) with 0.95 M HCl(aq). (a) before addition of any HCl (b) after addition of 25.0 mL of HCl (c) after addition of 50.0 mL of HCl (d) after addition of 75.0 mL of HCl (e) after addition of 100.0 mL of HCl

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.65....

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.65. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.85 M B(aq) with 0.85 M HCl(aq). a) before addition of any HCl b)after addition of 25.0mL of HCl c)after addition of 50.0mL of HCl d) after addition of 75.0mL of HCl e) after addition of 100.0 mL of HCl

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.79....

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.79. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.55 M B(aq) with 0.55 M HCl(aq). (a) before addition of any HCl (b) after addition of 25.0 mL of HCl (c) after addition of 50.0 mL of HCl (d) after addition of 75.0 mL of HCl (e) after addition of 100.0 mL of HCl

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.44....

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.44. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.80 M B(aq) with 0.80 M HCl(aq). a)before any HCl is added b)25ml HCl c)50ml HCl d)75ml HCl e)100ml HCl

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.58....

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.58. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.50 M B(aq) with 0.50 M HCl(aq).(a) before addition of any HCl (b) after addition of 25.0 mL of HCl (c) after addition of 50.0 mL of HCl (d) after addition of 75.0 mL of HCl (e) after addition of 100.0 mL of HCl

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