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# Acid base equilibria watch

1. I just finished learning how to find either pH of strong acid or concentration of H+ ions. However I'm stuck on a question.

Calculate the pH of the solution formed when 250 cm3 of water is added to 50cm3 of 0.200 moldm-3 HNO3.

I worked out the new H+ concentration to be 0.0333moldm-3.
2. -log10([H+])
3. Looks good so far.
What's the next step?
4. (Original post by TutorsChemistry)
Looks good so far.
What's the next step?
300/1000 * [H+] = 50/1000 x 0.200
[H+] = 0.0333 moldm-3 (monoprotic acid so no need to multiply anything?)

pH = -log10(0.0333)
= 1.48

Does this look right? I'm not confident I got the concentration bit right
5. That's what i got too.

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