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# Calculate the ph of CH3COOH of conc. 0.100 moldm-3

1. (Ka =1.74 *10-5 at 298k)

Conc. 0.100

1.74*10-5 =(H+)/(HA)= 0.1 moldm-3
(H+)= 1.319090569*10-3
10-1.31...=pH 0.996

???? Is this right
2. (Original post by Hazel99)
(Ka =1.74 *10-5 at 298k)

Conc. 0.100

1.74*10-5 =(H+)/(HA)= 0.1 moldm-3
(H+)= 1.319090569*10-3
10-1.31...=pH 0.996

???? Is this right
[H+]^2 = (Ka x [HA])
[H+] = square root of (Ka x [HA])
the square root of 1.74x10-5 x 0.100 = 1.319x10-3
-log(1.319x10-3) = 2.88

3. (Original post by Louisss;[url="tel:67768910")
67768910[/url]][H+]^2 = (Ka x [HA])
[H+] = square root of (Ka x [HA])
the square root of 1.74x10-5 x 0.100 = 1.319x10-3
-log(1.319x10-3) = 2.88

Thank you 😊

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