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

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SUMMARY

The % ionisation of hypochlorous acid (HClO) in a 0.015M solution at 25°C is calculated to be 0.14%. The acid dissociation constant (Ka) for HClO is 3.0 x 10^-8. The calculation involves setting up an equilibrium expression using the formula \(\frac{{[x][x]}}{{0.015 - x}} = Ka\), where x is the concentration of ionized HClO. The discussion highlights the use of ICE tables for solving equilibrium problems.

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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 celsius

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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