Asked by ruby
I am in the process of calculating the hydrogen ion concentration in a .25litre solution of sulfuric acid,.21g of sulfuric acid was disolved in water to make the solution. I have calculated the molarity of the acid to be 2.14x10^-3 mol per litre.
I also need help in calculating the pH of the acid soulution.
Well, if the H ion is twice that concentration..
pH= -log(2*2.14*10^-3)
grab your calculator.
I think you should check your calculations. I THINK you have 0.21/98 = 2.14 x 10^-3 mols but that is in 0.25 L. You want mols/L.
I have calculated the mols/L by multiplying .25ltrs and .21g by 4. This gives .84mols per litre. can this be confirmed for me please.
I also need help in calculating the pH of the acid soulution.
Well, if the H ion is twice that concentration..
pH= -log(2*2.14*10^-3)
grab your calculator.
I think you should check your calculations. I THINK you have 0.21/98 = 2.14 x 10^-3 mols but that is in 0.25 L. You want mols/L.
I have calculated the mols/L by multiplying .25ltrs and .21g by 4. This gives .84mols per litre. can this be confirmed for me please.
Answers
There are no human answers yet.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.