Quantitative Chemical Analysis Question, please give detailed instructions for e

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Quantitative Chemical Analysis Question, please give detailed instructions for each step so I can understand these types of questions.

5. (a). The pKw for water at 200oC is 11.289. What is the pH and [OH-] of water at 200o C?

(b). Write the reaction (equation) for the dissociation of acetic acid. Calculate the deltaGo of this reaction. Explain whether the reaction is spontaneous or not.

(c). Find the [H+] in 0.1M KCI. How does the [H+] value in 0.1M KCI, compare with that of [H+] in pure water? Give the percent increase or decrease.

Explanation / Answer

5.

a). The pKw for water at 200oC is 11.289. What is the pH and [OH-] of water at 200o C?

Kw = 10^-11.289 = 5.15*10^-12

Kw = [OH][H]

x*x = Kw

x = sqrt(Kw) = (5.15*10^-12)^0.5 = 2.26*10^-6

[OH-] = 2.26*10^-6

pH = -log(2.26*10^-6) =5.64

b). Write the reaction (equation) for the dissociation of acetic acid. Calculate the deltaGo of this reaction. Explain whether the reaction is spontaneous or not.

CH3COOH(aq) + H2O(l) <--> CH3COO-(aq) + H3O+(aQ)

dG° = -RTlnK

dG° = -8.314*298*ln(1.75 × 10 5) = 27137.6 kJ

This is not spotnanoeus (actually is an equilbiium)

c)

Find the [H+] in 0.1M KCI. How does the [H+] value in 0.1M KCI, compare with that of [H+] in pure water? Give the percent increase or decrease.

M = 0.1 KCl

Since KCl ---> K+ and Cl-

Cl- will not interact with H2O to form HCl, therefore [H+] ions in water are pretty similar to [H+] in the KCl soilution.

Expect something near [H+] = sqrt(Kw)

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