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aleksley [76]
3 years ago
7

In a 0.738 m solution, a weak acid is 12.5% dissociated. calculate ka of the acid.

Chemistry
1 answer:
luda_lava [24]3 years ago
7 0
<h3><u>Answer;</u></h3>

Ka == 1.153 × 10^-4

<h3><u>Explanation</u>;</h3>

Weak acid is HA;

Equilibrium: HA(aq) + H2O(l) <=> H3O^+(aq) + A^-(aq)

Ka = [H3O^+][A^-]/[HA]  

[H3O^+] = [A^-] = (0.0125)(0.738)

                         = 0.00923 M and [HA]

                         = 0.738 M - 0.00923 M = 0.738 M because the value 0.00923  is insignificant in this case.  

Ka = (0.00923)^2/(0.738)

     <u>= 1.153 × 10^-4</u>

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diamong [38]
Hey there! 

<span>Use the equation of Clapeyron:
</span>
T in kelvin :

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P *V = n * R * T

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5 0
4 years ago
Isopropyl alcohol is mixed with water to produce a 32.0 % (v/v) alcohol solution. How many milliliters of each component are pre
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<u>Answer:</u> The volume of isopropyl alcohol present is 254.4 mL and the volume of water present is 540.6 mL

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