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4.84
pH + pOH = 14. So pOH = 14 - 10.95 = 3.05 pOH = -log[OH-] [OH-] = 8.91 x 10-4 M
The atomic weight of Na-OH is 23+16+1= 40. So to prepare the one N solution of Na-OH, you need to add 40 grams of Na-OH in one litre of water. To prepare the 4N Na-OH solution, you need to add 160 grams of Na-OH in one litre of water.
A solution with H+ = 1.2 x 10-4 a solution with OH- = 2.5 x 10 -9 A solution with pH = 4.5
2(OH) + Mg ------> Mg(OH)2
solution with [OH-] = 2.5 x 10-9 , A solution with [H+] = 1.2 x 10-4, A solution with pH = 4.5
4.84
pH + pOH = 14. So pOH = 14 - 10.95 = 3.05 pOH = -log[OH-] [OH-] = 8.91 x 10-4 M
The atomic weight of Na-OH is 23+16+1= 40. So to prepare the one N solution of Na-OH, you need to add 40 grams of Na-OH in one litre of water. To prepare the 4N Na-OH solution, you need to add 160 grams of Na-OH in one litre of water.
A solution with H+ = 1.2 x 10-4 a solution with OH- = 2.5 x 10 -9 A solution with pH = 4.5
It increases the concentration of OH - in a solution.
Oh my gollywogs
[H+] = Kw / [OH-] = 1.0*10-14 / 2.5*10-4 = 4.0*10-11 mol/L
-log[1 X 10^-4 M OH(-)] = 4 14 - 4 = 10 pH ----------------
2(OH) + Mg ------> Mg(OH)2
No. If the number of H+ and OH- ions are equal then the solution is neutral. A solution is considered alkaline if it has more OH- ions than H+ ions.
A base in solution will produce hydroxide or OH- ions.