To make a solution as pH 10 how much ammonium hydroxide to be added?

Question

(if using deionised water volume as 1m3)

Answer 1

Good God people!!!

pH is -log[H+] so [H+]=10^-10 and [OH-]=10-4 mole/lt

How could you possibly think 10 billion mole/lt is correct?????

Also ammonium hydroxide is a solution of 12-44% ammonia in water and it is not a strong electrolyte pKb=4.75

so Kb=[NH4+][OH-]/[NH4OH]=10^-pKb

From the stoichiometry you have [NH4+]=[OH-] and if Cis the total concentration of NH4OH you have

10^-4* 10^-4/ (C-10^-4)=10^-4.75=>
C=(10^-8+10^-8.75)/10^-4.75=6.62 *10^-4 mole/lt

If you want a final volume of 1m3=1000 liters you need 6.62 * 10^-1 mole. You don't give us the concentration of the original stock solution so I can't tell you to how much volume this corresponds to.

Answer 2

A ph of 10 corresponds with a concentration of H+-ions of 10^-10:
[H+]=10^-10
Assuming the water is 25 degrees Centrigade the pKw of water is 10^-14
pKw=[H+] x [OH-]=10^-14
That means the concentration of OH- is 10^-4 mol/liter.

Ammonium hydroxide forms an equilibrium with dissolved ammonia

NH3 + H20 <---> NH4+ + OH-

Kb = [NH4+]*[OH-] / [NH3] = 1.8*10^-5

Since the concentration of ammonium ions is equal to the concetration of hydroxide ions [NH4+]=[OH-], the above equation can be solved:

[NH3]=[NH4+]*[OH-]/Kb = [OH-]^2/Kb = 10^-8/1.8^-5 = 5.6*10^-4 mol/l

Because all reactions are equimolar the total of ammoniumhydroxide that has to be dissolved is equal the sum of the ammonium and ammonia at the given equilibrium.

Total to dissolve= [NH4+] + [NH3] = 10^-4 + 5.6*10^-4 = 6.6*10 ^-4 mol/l

Given the volume of 1m3 (=1000 liters) the total number of moles to be dissolved is 6.6*10^-4*1000= 0.66 mol or 0.23 grams.

Answer 3

10^4 mol dm^3

Answer 4

pH=log[H+]
so [H+]=10^10
Kw=[H+][OH-]=1x10^14
so [OH-]=10^14/10^10 =10^4mol dm^-3