Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: rwooduk1 on February 19, 2020, 01:50:47 PM
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I have a copper (II) sulphate, lactic acid, nitric acid solution (100 mL) with pH 2.3. I want to get this solution to pH 12, but I don't want to significantly increase the amount of solution. If say, I wanted to use 10 mL of NaOH to raise the pH to pH 12 what concentration NaOH would I need? I'm assuming the concentration would need to be extremely high, would the amount of NaOH needed even dissolve in 10 mL of DI? Is there a simple way to make this calculation?
If someone could point me in the right direction it would be appreciated.
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Because of this mixture it is better to know the amounts of the acids and the copper sulfate. These you have to neutralize first and additionally to calculate how much NaOH correspond to pH 12.
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Its highly soluble, try adding a drop of 1N and check pH, if no real change add a drop of 5N and so on.
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Thank you for the replies.
Copper sulfate 0.15 M, lactic acid 3 M and 10% nitric acid.
As suggested, I will simply trial and test with higher concs of NaOH.
Thanks again.
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Your information is not reliable (pH=2.3). 10 % nitric acid shows a pH slightly below 0.
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Interesting, that's the reading the pH meter gave, I didn't even know pH could go below zero to be honest.
I think I made a mistake with my calculation for the nitric acid, if the acid solution is 70% HNO3 and I want 10% HNO3 solution in my 200 mL Cu.C3H6O3 solution, how much should I be using?
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To calculate anything you need to have the right data set.
If you use solution volumes, you should have all reagent concentrations in molar concentrations. Then the calculations are quite simple.
If you express concentrations in a mixed way (molar and %), it is very difficult to find concentrations for such a mixture in uniform units. 10% HNO3 corresponds to a concentration of 1.7 M approximately.
Assuming you have 100 mL of mixed solution with concentrations: 0.15 M CuSO4, 3 M of lactic acid and 1.7 M HNO3 you need 0.5 mol of NaOH for neutralization reaction that gives you pH slightly above 8. Maximum concentrated NaOH may have a concentration of about 18 M at RT which corresponds to the NaOH content in a 10 mL solution of 0.18 mol and this means that to neutralize your solution you need about 30 mL of the most concentrated NaOH or about 10 mL of solid NaOH. These estimates indicate that your data is unrealistic or your solution whose pH you have measured has completely different concentrations.
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To calculate anything you need to have the right data set.
If you use solution volumes, you should have all reagent concentrations in molar concentrations. Then the calculations are quite simple.
If you express concentrations in a mixed way (molar and %), it is very difficult to find concentrations for such a mixture in uniform units. 10% HNO3 corresponds to a concentration of 1.7 M approximately.
Assuming you have 100 mL of mixed solution with concentrations: 0.15 M CuSO4, 3 M of lactic acid and 1.7 M HNO3 you need 0.5 mol of NaOH for neutralization reaction that gives you pH slightly above 8. Maximum concentrated NaOH may have a concentration of about 18 M at RT which corresponds to the NaOH content in a 10 mL solution of 0.18 mol and this means that to neutralize your solution you need about 30 mL of the most concentrated NaOH or about 10 mL of solid NaOH. These estimates indicate that your data is unrealistic or your solution whose pH you have measured has completely different concentrations.
Thank you AWK, this is extremely helpful. I will recheck my concentrations, it seems something has gone wrong somewhere.