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# Dilution Effect

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1. So for one of the OCR salters B chemistry past papers, there's a question about mixing HNO3 with NaOH.

20cm^3 of 0.015moldm^-3 HNO3 are mixed with 10cm^3 of 0.015moldm^-3 NaOH. Calculate the pH of the resulting solution.

The mark scheme isn't the clearest with this question, in the sense that I don't know what the concentration of H+ ions is.
It's question 4)d)v). http://www.ocr.org.uk/Images/144773-...-by-design.pdf

And here's the mark scheme!

http://www.ocr.org.uk/Images/142262-...gn-january.pdf
2. (Original post by Chubbiepanda)
So for one of the OCR salters B chemistry past papers, there's a question about mixing HNO3 with NaOH.

20cm^3 of 0.015moldm^-3 HNO3 are mixed with 10cm^3 of 0.015moldm^-3 NaOH. Calculate the pH of the resulting solution.

The mark scheme isn't the clearest with this question, in the sense that I don't know what the concentration of H+ ions is.
It's question 4)d)v). http://www.ocr.org.uk/Images/144773-...-by-design.pdf

And here's the mark scheme!

http://www.ocr.org.uk/Images/142262-...gn-january.pdf

Formula of pH is pH = -log [H+] so pH = -log [0.015] of HNO3 as HNO3 is a weak acid so dissociates into HNO3 ---> H+ + NO3- as the H+ will be reacting with the dissociation of NaOH so it is negligible.
3. (Original post by Gav107)
Formula of pH is pH = -log [H+] so pH = -log [0.015] of HNO3 as HNO3 is a weak acid so dissociates into HNO3 ---> H+ + NO3- as the H+ will be reacting with the dissociation of NaOH so it is negligible.
If I use 0.015moldm^-3 I get pH = -log [0.015] = 1.82
But in the answer they get pH = 2.30 ???
4. (Original post by Chubbiepanda)
If I use 0.015moldm^-3 I get pH = -log [0.015] = 1.82
But in the answer they get pH = 2.30 ???
oh sorry looked at the wrong answer i did the other part i will take a look in to that, sorry!

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