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Chemistry help. Thanks in advance.

If 17.52ml of a 0.519 mol dm^-3 solution of NaOH is titrated with a 0.582 mol dm^-3 solution of nitric acid, what volume is required to reach the equivalent point?

The answer is 7.82ml in my textbook.

When I use MaVa=MbVb I keep getting 15.62ml. Please help me!
Original post by xpointx
If 17.52ml of a 0.519 mol dm^-3 solution of NaOH is titrated with a 0.582 mol dm^-3 solution of nitric acid, what volume is required to reach the equivalent point?

The answer is 7.82ml in my textbook.

When I use MaVa=MbVb I keep getting 15.62ml. Please help me!


I got the same answer as you got!

NaOH + HNO3 --> NaNO3 + H2O
n(NaOH) = 0.519 x 0.01752 = 0.00909288 = n(HNO3)

V(HNO3) = 0.00909288 / 0.582 = 0.01562 L = 15.62 mL


Posted from TSR Mobile
The text book is wrong and the answer is 15.62.
It may have been changed from sulphuric acid to nitric acid for some reason.
Text books are often riddled with these sorts of errors.
Original post by Madasahatter
The text book is wrong and the answer is 15.62.
It may have been changed from sulphuric acid to nitric acid for some reason.
Text books are often riddled with these sorts of errors.


Thanks for that, I couldn't figure out where I went wrong! :smile:


Posted from TSR Mobile

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