Finding pH and molarities

  • Thread starter Clari
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  • #1
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Assuming the first ionization of sulphuric acid is 100% and the [tex]K_{a2}[/tex] of sulphuric acid is 0.10M at 25 degree celsius.(Ka is the acid disocciation constant)
a. Find the pH of a 0.1M [tex]NaHSO_4[/tex]
b. Calculate the molarities of [tex]H^+[/tex], [tex]HSO_4^-[/tex], [tex]SO_4^{2-}[/tex] respectively in 0.1M [tex]H_2SO_4[/tex]

Here are my steps:
a. Let a be the degree of dissociation.
[tex]K_{a2} = \frac{a^2}{0.1-a} = 0.1[/tex]
a = 0.0618
pH = - log 0.0618 = 1.21

I have no idea of how to do part b. Please help me with it and tell me whether I have done right in part a.
 

Answers and Replies

  • #2
GCT
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I've already explained this problem in multiple threads, nevertheless,

a)

[tex]Ka_2= \frac{[SO_4^{-2}][H_3O^+]}{[HSO_4^-]} [/tex]

[tex]Ka_2= \frac{[x][x]}{[.1M-x]} [/tex]

b) The first acid will dissociate completely, thus the initial concentration of both the HSO4- and H3O+ will be .1M. Thus

[tex]Ka_2= \frac{[SO_4^{-2}][H_3O^+]}{[HSO_4^-]} [/tex]

[tex]Ka_2= \frac{[x][.1M+x]}{[.1M-x]} [/tex]
 

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