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 12M HCl (about 6.2ml) and dilute to 15ml with deionized water.
When about 6ml of 12M HCl was added, a white ppt was obtained.
What happened?
Thanks.
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
You have now specified 12 M HCl, that is not dilute.
The white ppt was NaCl not in solution. You added a lot of Cl ions with the 12M acid, you already had Na ions. So many, they could not go into solution.
To perform the limit test for sulphates in an NaOH sample, follow these steps:
1. Dissolve 3.0g of NaOH in 6ml of deionized water.
- This means you need to measure 3.0g of NaOH using a balance and add it to a clean container. Then add 6ml of deionized water into the container and stir well until the NaOH is completely dissolved.
2. Adjust the pH of the solution to 7 with 12M HCl.
- Take 12M HCl solution and slowly add it to the NaOH solution while stirring, until the pH of the solution reaches 7. This can be done by using a pH meter or pH indicator paper.
3. Dilute the solution to 15ml with deionized water.
- After reaching pH 7, add deionized water to the solution until the total volume reaches 15ml. This step ensures that the concentration of the solution is appropriate for the test.
When about 6ml of 12M HCl was added, a white precipitate (ppt) was obtained.
- This means that after adding approximately 6ml of 12M HCl to the NaOH solution, a white solid formed in the solution. The white precipitate indicates the presence of sulphates in the NaOH sample.
Concentration of HCl used:
- In the given procedure, 12M HCl is used. This means that the concentration of the HCl solution is 12 moles per liter. The volume of HCl needed to neutralize the NaOH can be determined by calculating the moles of NaOH and using stoichiometry. In this case, a molarity of approximately 10M is needed, and 32% HCl is around 10M concentration.