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Pavlova-9 [17]
3 years ago
6

Nh3 is a weak base (kb = 1.8 × 10–5 m) and so the salt nh4cl acts as a weak acid. what is the ph of a solution that is 0.013 m i

n nh4cl?
Chemistry
1 answer:
ankoles [38]3 years ago
6 0
First, we need to get the value of Ka:

when Ka = Kw / Kb 

we have Kb = 1.8 x 10^-5 

and Kw = 3.99 x 10^-16  so, by substitution:

Ka = (3.99 x 10^-16) /  (1.8 x 10^-5) = 2.2 x 10^-11

by using the ICE table :

                 NH4+  + H2O →NH3 +  H+
intial          0.013                       0         0 

change       -X                          +X      +X

Equ        (0.013-X)                      X         X

when Ka = [NH3][H+] / [NH4+] 

by substitution:

2.2 x 10^-11 = X^2 / (0.013 - X)  by solving this equation for X

∴X = 5.35 x 10^-7

∴[H+] = X = 5.35 x 10^-7

∴PH = - ㏒[H+]
 
        = -㏒(5.35 x 10^-7)
        = 6.27
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Explanation:

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             Rate = k[HgCl_{2}]x [C_{2}O^{2-}_{4}]y[/tex]

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       1.8 \times 10^{-5} = k[0.105]x [0.15]y ............. (1)

       7.2 \times 10^{-5} = k [0.105]x [0.30]y ........... (2)

       3.6 \times 10^{-5} = k [0.0525]x [0.30]y ............ (3)

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            1.8 \times 10^{-5} = k [0.105]1 [0.15]2  

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