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Final pH in tank

Question asked by Bjarne Gullov-Rasmussen on Jan 22, 2014
Latest reply on Jan 23, 2014 by Guest

Exercise:

I need a final pH of 9 in a tank, obtained by disolving NaOH in water.

Solution, in part:
Molar mass af NaOH is 40 gram.

pH++pOH- = 14 thus  pOH- = 5

M = molarity = mol/liter

1 mol = 6,2 x 1023

 

So:

5 = pOH- = log10-5     = 0,0001 mol/liter

By a molar mass of 40 g/liter a pH of 9 should be 40 x 0,0001 = 0,004 g/liter of NaOH

 

That does not look right to me, but what is wrong ?

rgdr´s

bjarne.gullovrasmussen@alfalaval.com

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