What is the % ionisation of hypochlorous acid in a 0.015M solution at 25°C?

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The discussion focuses on calculating the percentage ionization of hypochlorous acid (HClO) in a 0.015M solution at 25°C, given its dissociation constant (Ka) of 3.0 x 10^-8. The solution involves setting up an equilibrium expression where the concentration of ionized HClO is represented by 'x'. Participants mention using the ICE table method to organize the concentrations of reactants and products. The final calculated percentage ionization is determined to be 0.14%. The conversation highlights the importance of understanding weak acid dissociation in aqueous solutions.
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Homework Statement


The Ka of hypochlorous acid (HClO) is 3.0*10^-8 at 25 degrees C. What is the % ionisation of hypochlorous acid in a 0.015M aqueous solution of HClO at 25 degrees Celcius

The answer is 0.14

Homework Equations


-log[H+], n = cV, pH = pKa + log (base/acid), x/y*100=%


The Attempt at a Solution


I tried -log3*10^-8 and got 7.5. Not to sure what to do with that.
 
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Was the "0.015" actually the FORMAL concentration of the acid? If we assume that the concentration really was meant as 0.015M of the weak acid, then

\[<br /> \frac{{\left[ x \right]\left[ x \right]}}{{0.015 - x}} = Ka = 3 \times 10^{ - 8} <br /> \]<br />, where x represents approximately the amount of moles per liter of HClO which actually ionizes.
 
NOP90, have you heard about ICE tables?
 
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