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A Level Chemistry Question help - Acids and Bases

A student carried out an investigation with aqueous solutions of nitric acid, sodium hydroxide, ethanoic acid, and water.
The student diluted 0.015 mol dm^–3 nitric acid with an equal volume of water and measured the pH of the diluted acid at 25 °C.
(i) Calculate the pH of 0.015 mol dm^–3 nitric acid. [2]
(ii) Calculate the pH of the diluted acid. [1]


I have done part (i) successfully however I don't know how to do part (ii). Can someone please help? Thank you.
Screenshot 2021-08-16 at 20.25.39.png
here's the full question for anyone who needs it!
By diluting the acid in an equal volume of water, you have halved the concentration (if you have volume V at the start this is then 2V at the end).
Original post by EierVonSatan
By diluting the acid in an equal volume of water, you have halved the concentration (if you have volume V at the start this is then 2V at the end).

It seems right as that's how they got to the answer but I don't understand it. If you don't mind could you explain what you just said in a detailed way, please? thank you :smile:
Original post by jeebie_3010
It seems right as that's how they got to the answer but I don't understand it. If you don't mind could you explain what you just said in a detailed way, please? thank you :smile:

How about a simple way? :tongue:

If you have 100 cm3 of acid and add another 100 cm3 of water, you end up with ... 200 cm3

This means you have doubled the volume, which is the same as halving the concentration. (This is because molar concentration = moles / volume)

It doesn't matter what volume there is at the start: 10, 20, 50 , 1000 etc
(edited 2 years ago)
Original post by EierVonSatan
How about a simple way? :tongue:

If you have 100 cm3 of acid and add another 100 cm3 of water, you end up with ... 200 cm3

This means you have doubled the volume, which is the same as halving the concentration. (This is because molar concentration = moles / volume)

It doesn't matter what volume there is at the start: 10, 20, 50 , 1000 etc

Ah okay, I understand it better now, I was finding it hard to visualise the problem!
Thank you so much for your help! :smile:
Original post by jeebie_3010
Ah okay, I understand it better now, I was finding it hard to visualise the problem!
Thank you so much for your help! :smile:

Welcome :h:

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