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Preparing a solution of hydroxylamine hydrochloride in HCl, struggling
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Judas Matete
Preparing a solution of hydroxylamine hydrochloride in HCl, struggling
Does it imply that the 2mL of solution is made up of equal parts (1mL) of 0.25M hydroxylamine HCl and 0.25M HCl?
No, if you mixed 1 mL of 0.25M HCl and 1 mL of hydroxylamine hydrochloride, you'd have 2 mL of a solution that was 0.125M in HCl and 0.125M in hydroxylamine hydrochloride.
I think the purpose of the hydrochloric acid is to ensure that the pH stays acid. One way you could make this solution is by:
Decide on a volume of solution you want to use, say $X$ mL.
Prepare at least $X$ mL of 0.25M hydrochloric acid by dilution from a more concentrated stock solution.
Weigh out the amount of $\ce{NH2OH.HCl}$ that you need. It has a formula weight of 69.49 g/mol, so for $X$ mL of a 0.25 M solution you will need:
Dissolve the weighed out powder with your 0.25M $\ce{HCl}$ solution in a volumetric flask, first fully dissolving the powder and then slowly adding more HCl until the target volume is reached.
The resulting solution will be 0.25M in both hydrochloric acid and in hydroyxlamine hydrochloride.
Does it imply that the 2mL of solution is made up of equal parts (1mL) of 0.25M hydroxylamine HCl and 0.25M HCl?
No, if you mixed 1 mL of 0.25M HCl and 1 mL of hydroxylamine hydrochloride, you'd have 2 mL of a solution that was 0.125M in HCl and 0.125M in hydroxylamine hydrochloride.
I think the purpose of the hydrochloric acid is to ensure that the pH stays acid. One way you could make this solution is by:
Decide on a volume of solution you want to use, say $X$ mL.
Prepare at least $X$ mL of 0.25M hydrochloric acid by dilution from a more concentrated stock solution.
Weigh out the amount of $\ce{NH2OH.HCl}$ that you need. It has a formula weight of 69.49 g/mol, so for $X$ mL of a 0.25 M solution you will need:
Dissolve the weighed out powder with your 0.25M $\ce{HCl}$ solution in a volumetric flask, first fully dissolving the powder and then slowly adding more HCl until the target volume is reached.
The resulting solution will be 0.25M in both hydrochloric acid and in hydroyxlamine hydrochloride.
No, if you mixed 1 mL of 0.25M HCl and 1 mL of hydroxylamine hydrochloride, you'd have 2 mL of a solution that was 0.125M in HCl and 0.125M in hydroxylamine hydrochloride.
I think the purpose of the hydrochloric acid is to ensure that the pH stays acid. One way you could make this solution is by:
$$0.25 \frac{\mathrm{mol}}{\mathrm{L}} \times \frac{69.49\mathrm{g}}{\mathrm{mol}} \times \frac{X\;\mathrm{mL}}{1} \times \frac{\mathrm{L}}{\mathrm{1000mL}}=\frac{69.49\;X}{4000}\mathrm{g\;of\;\ce{NH2OH.HCl}}$$
The resulting solution will be 0.25M in both hydrochloric acid and in hydroyxlamine hydrochloride.
No, if you mixed 1 mL of 0.25M HCl and 1 mL of hydroxylamine hydrochloride, you'd have 2 mL of a solution that was 0.125M in HCl and 0.125M in hydroxylamine hydrochloride.
I think the purpose of the hydrochloric acid is to ensure that the pH stays acid. One way you could make this solution is by:
$$0.25 \frac{\mathrm{mol}}{\mathrm{L}} \times \frac{69.49\mathrm{g}}{\mathrm{mol}} \times \frac{X\;\mathrm{mL}}{1} \times \frac{\mathrm{L}}{\mathrm{1000mL}}=\frac{69.49\;X}{4000}\mathrm{g\;of\;\ce{NH2OH.HCl}}$$
The resulting solution will be 0.25M in both hydrochloric acid and in hydroyxlamine hydrochloride.
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