Dilute commercial concentrated ammonium hydroxide with water to a concentration of your choice.

0.3%NH4OH means 0.176M NH3 solution in 100mlH2O

M1V1 = M2V2

2*V1 = 0.176*100 (If you need 100 ml solution)

V1 = ml of 2M NH3 in MeOH (take the V1 and mix with H2O to make 100 ml standard solution)

for this you must be  dilute commercial concentrated ammonium hydroxide with distilled water to a concentration  as per your requirement.

Thanks a lot

shall we use the diluted ammonia solution as ammonia hydroxide solution?

Regardin this question I am confused about one thing: For example, if an eluent for solid phase extraction consists of 5% NH4OH in methanol, does it mean that I simply have to dilute for example a commercially available concentrated aqueous 25% NH4OH 5-times with methanol or I already have to start with concentrated methanolic solution of NH4OH and then dilute it accordingly with methanol?

Baskaran Subbian's question. In fact don't be confused. Diluted ammonia solution and ammonia hydroxide solution is not different. If i understand Matej Somrak very well, diluted ammonia and ammonium hydroxide are same. The diluted ammonia is the commercial ammonia of about 28 - 30% by volume by addition of solution which contains hydroxyl example water. Ammonia itself contains only two atoms (H = 3 and N = 1) = NH3 Then when it reacts with water it become ammonium hydroxide (NH4OH).

However, since ammonia is a weak base and methanol is much less acidic that water, there is for all practical purposes no reaction. If it’s aqueous ammonia solution and methanol, you get a 3-component mixture (with small amounts of ammonium hydroxide, already present in the aqueous ammonia), but again, effectively no reaction. Barry Gehm (7th July, 2018). in his response to such question "What happens if you mix ammonia with methanol"? Quora. I think this will help you understand more about you question.

i) We should generally take not fully specified % concentrations as wt% ― where wt is often omitted or can be substituted by either wt., wt/wt, wt./wt., or m/m ― meaning: (mass of pure solute / mass of solution)·100%; even if this is not clearly stated. If is required to prepare the requested solution to exactly fullfil a given volume, the density of the solution for that same concentration, at a given temperature, should be known. Concentrations are typically given in a solute to solution basis; not in a solute to solvent basis. A rare exception is molality (mol solute / kg solvent).

ii) The initial (conc.) sol. is in principle to be taken as 2.00 M in NH3. I shall admit that the concentration of the final (dil.) sol. is 0.3 wt% (solute:solution); so that the indicated unspecified 0.3% does not refer to 0.3 % v/v dilution of the initial solution (vol. initial solution : vol. final solution). Also note that 0.3 % seems too low for a typical dilution ratio.

iii) Note that for the initial (conc.) sol.: (2.00 mol NH3/dm3)·(17.03 g NH3/mol NH3)/{(1000 cm3/dm3)·0.787(g methanolic sol. / cm3)} = 2.00·(17.03 g NH3)/{103·0.787(g methanolic sol.)} = 4.33·10-2 g NH3/ g methanolic sol. = 4.33·10-2·100 wt% = 4.33 wt% (solute:solution).

iv) Mass balance for the solute (NH3) on the dilution operation:

{4.33 wt% (solute:solution)}·minitial solution =

{0.3 wt% (solute:solution)}·mfinal solution

v) Mass balance for the solution on the dilution operation:

minitial solution + madded solvent = mfinal solution

vi) Example: for mfinal solution = 100.00 g; minitial solution = {{0.3 wt% (solute:solution)}·mfinal solution}/{4.33 wt% (solute:solution)} = 6.93 g; madded solvent = 100.00 - 6.93 = 93.07 g. Hence, we can take 6.93 g of 4.33 wt% NH3 methanolic solution, to be mixed with 93.07 g of solvent (methanol), to obtain 100 g of the intended 0.3 wt% dilute sol.