Why 7ml of 10M HCl? Do I add 7ml of 10M HCl 1st slowly in ice bath, then slowly adjust the pH to 7 using 10M HCl? Then to dilute to 15ml in a graduated cylinder?
Thanks
Posted by candy on Tuesday, September 21, 2010 at 10:03am.
The following is part of a procedure for a limit test for sulphates in an NaOH sample:-
Dissolve 3.0g of NaOH in 6ml of deionized water, adjust to pH 7 with HCl (approx.7.5ml) and dilute to 15ml with deionized water.
Please explain how to perform this procedure. Conc HCl or 1M HCl? I do not understand it.
Thanks.
Chemistry - urgent - bobpursley, Tuesday, September 21, 2010 at 10:09am
What form of NaOH did you use? Assume granular.
moles NaOH=3/40
moles HCl needed= 3/40=volume*Molarity
so lets see what 7 ml will neutralize
molarity=3/40/.007= or about 10Molar. 32Percent HCl is about 10M
To perform the procedure for the limit test for sulfates in an NaOH sample, you will need to follow these steps:
1. Dissolve 3.0g of NaOH in 6ml of deionized water.
- This means you need to mix 3.0g of solid NaOH with 6ml of deionized water until the NaOH is fully dissolved.
2. Adjust the pH to 7 with HCl.
- In this step, you need to use hydrochloric acid (HCl) to adjust the pH of the solution to 7. The amount of HCl needed can be calculated based on the stoichiometry of the reaction.
- First, calculate the moles of NaOH used:
Moles of NaOH = mass of NaOH / molar mass of NaOH = 3.0g / 40g/mol ≈ 0.075 mol
- The balanced chemical equation for the reaction between NaOH and HCl is:
NaOH + HCl → NaCl + H2O
- From the equation, you can see that 1 mole of NaOH reacts with 1 mole of HCl. Therefore, the moles of HCl needed will be the same as the moles of NaOH used, which is approximately 0.075 mol.
- Now, you can calculate the volume of 10M HCl needed:
Volume of 10M HCl = Moles of HCl needed / Molarity of HCl = 0.075 mol / 10 mol/L = 0.0075 L = 7.5ml
- So, you will need approximately 7.5ml of 10M HCl to adjust the pH to 7.
3. Dilute to 15ml with deionized water.
- After adjusting the pH with HCl, you need to add enough deionized water to bring the final volume up to 15ml. This can be done using a graduated cylinder to accurately measure the amount of water added.
It is important to note that the concentration of the HCl solution is not explicitly provided in your question. However, based on the calculations above, it is assumed that a 10M HCl solution is used. If you have a different concentration of HCl, you will need to adjust the calculations accordingly.