Chemistry, 18.02.2020 05:22, Mangolinux7173
Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid at 25 oC. b) Determine the percent dissociation for the solution.
Answers: 3
of acetic acid at 
is 
= ?![pH=-log[H^+]](/tpl/images/0513/9443/15713.png)
![2.87=-log[H^+]](/tpl/images/0513/9443/3a07c.png)
![[H^+]=1.35\times 10^{-3}M](/tpl/images/0513/9443/01bcb.png)
![[CH_3COO^-]=1.35\times 10^{-3}M](/tpl/images/0513/9443/0a4ad.png)
![[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M](/tpl/images/0513/9443/5420f.png)
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Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calc...
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