Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: whatheck on March 04, 2011, 09:04:46 PM
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calculate the pH value of 0.1mole CH3COONa in water. Ka of CH3COOH = 1.8x10^-5, Kw = 1x10^-14
The equation should be CH3COONa + H2O <---> CH3COOH + NaOH, right ?
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i have calculated for several times and got three values, 8.26,9.26 & 11.1. Which one is correct or not above all?
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Show how you got these values.
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BY pH = pK + log[salt]/[acid], putting all those numerber inside will get a <pH 7, but that should be an alkaine value. I think the problem should be the pK but anyway i don't know.
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putting all those numerber inside
Which "all" numbers?
Hint: you have a solution of conjugate base
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i think it should be taken as the case of salt hydrolysis of a salt of weak acid and strong base and formula applied is
pH = 7+1/2(pka) + 1/2(logc)
where pka = -log ka = 4.74
log c = log (0.1) (assuming the volume of water is 1 litre.
so answer should be 8.87 (approx)
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pH = 7+1/2(pka) + 1/2(logc)
[OH-]=SQRT(Kbxc)
[H3O+]=Kw/[OH-]=Kw/SQRT(Kbxc)=SQRT(Kw2/(Kbxc))
Hence
[H3O+]=SQRT(KwxKa/c)
or in logarithmic scale
p[H]=1/2(pKw+pKa -log(c)) where p[H] means -log from concentration of [H3O+]