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Hello could anyone help me answer this and could you number the steps? Watch

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1. Calculate the ph at 298k of an aq solution of CH3COOH of concertration 0.100 mol dm-3 . (Ka = 1.74*10-5 mol dm-3 at 298k)
3. (Original post by Hazel99)
Calculate the ph at 298k of an aq solution of CH3OOH of concertration 0.100 mol dm-3 . (Ka = 1.74*10-5 mol dm-3 at 298k)
4. (Original post by Daniel Atieh)
I don't know if this is right but I had a go.

Ka= 1.74*10-5
Conc= 0.1 M

1.74*10-5 = (H+) / (HA) = 0.1 mol dm-3

(H+)= 1.319090569*10-3

10- 1.31..... = ph 0.996

5. (Original post by Hazel99)
Calculate the ph at 298k of an aq solution of CH3OOH of concertration 0.100 mol dm-3 . (Ka = 1.74*10-5 mol dm-3 at 298k)
Methyl hydroperoxide?
6. (Original post by alow)
Methyl hydroperoxide?
Seems unlikely. There's likely a missing C.
7. (Original post by Kvothe the Arcane)
Seems unlikely. There's likely a missing C.
Yeah, I think I remember the Ka of ethanoic acid is about 1.8*10-5 too.
8. Deffo EtOOH, I always use 1.74 in class.
9. (Original post by alow;[url="tel:67756552")
67756552[/url]]Methyl hydroperoxide?
Are u trying to name CH3OOH??
10. (Original post by Hazel99)
Are u trying to name CH3OOH??
Yeah that's the name of CH3OOH. You've got the formula wrong, you mean CH3COOH.
11. (Original post by alow;[url="tel:67757604")
67757604[/url]]Yeah that's the name of CH3OOH. You've got the formula wrong, you mean CH3COOH.

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