Posted by Stuck on Saturday, May 15, 2010 at 12:16am.
"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).

Chemistry  DrBob222, Saturday, May 15, 2010 at 1:21pm
nHNO3: (0.225 mol/l)(12.6 mL)/1000
= 2.835 x 10^3 mol
This step is ok.
nNa3BO3: (2.835 x 10^3 mol)(1/3)
= 9.45 x 10^4 mol
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
[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).
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.
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