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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