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
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 ?