Question

"An aqueous sodium borate solution is titrated to the second endpoint with 0.225 mol/l nitric acid in the presence of a suitable indicator. If an average of 12.6mL of titrant are required to reach the second endpoint for a 25.0 mL sample, what is the base concentration?"

My calculations:

nHNO3: (0.225 mol/l)(12.6 mL)/1000
= 2.835 x 10^-3 mol

nNa3BO3: (2.835 x 10^-3 mol)(1/3)
= 9.45 x 10^-4 mol

[Na3BO3]: (9.45 x 10^-4 mol)/(25 mL)(1000)
= 0.0378 mol/L

Is this correct? Because the answer given is different (0.0567 mol/L).

Answer 1

nHNO3: (0.225 mol/l)(12.6 mL)/1000
= 2.835 x 10^-3 mol
<b>This step is ok.</b>

nNa3BO3: (2.835 x 10^-3 mol)(1/3)
= 9.45 x 10^-4 mol
<b>This step is in error. The problem says that Na3BO3 was titrated to the second end point (meaning two H ions were added) but you divided by 3. You should have divided by 2</b>

[Na3BO3]: (9.45 x 10^-4 mol)/(25 mL)(1000)
= 0.0378 mol/L

Is this correct? Because the answer given is different (0.0567 mol/L).
<b>Another way to approach the problem is to say if 12.6 mL were used to add 2 Hs, then 6.3 would have been used to add 1 H, then go through the remainder of the calculation. I think the answer of 0.0567 M is correct. </b>