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radusevciuc7719
12.03.2020 •
Chemistry
If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH of the titration mixture? (For HC2H3O2 , Ka = 1.82 x 10-5)
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Ответ:
8.73
Explanation:
The concentration of acetic acid can be determined as follows:
Moles of
= ![95.27* 10^{-3}* 0.094](/tpl/images/0545/6807/96ee5.png)
=0.0090 moles
Moles of![KOH = 79.06*10^{-3}*0.113](/tpl/images/0545/6807/3f159.png)
= 0.0090 moles
The equation for the reaction can be expressed as :
Concentration of
ion = ![\frac{0.0090}{Total volume (L)}](/tpl/images/0545/6807/f98fc.png)
=![\frac{0.0090}{(95.27+79.06)} *1000](/tpl/images/0545/6807/a4fea.png)
= 0.052 M
Hydrolysis of
ion:
⇒![\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}](/tpl/images/0545/6807/f2627.png)
=![0.5494*10^{-9}= \frac{x*x}{0.052-x}](/tpl/images/0545/6807/fbfdc.png)
As K is so less, then x appears to be a very infinitesimal small number
0.052-x ≅ x
pH = 14 - pOH
pH = 14 - 5.27
pH = 8.73
Hence, the pH of the titration mixture = 8.73
Ответ:
Explanation:
A homomonocyclic compound composed of eight sulfur atoms. ...