a solution is initially 0.0884 M in an acid which is 21.4% ionized at equilibrium. Calculate...

a solution is initially 0.0884 M in an acid which is 21.4% ionized at equilibrium. Calculate the Ka for the solution at 25C (assume all acids are monoprotic)

Expert Solution

Let the given monoprotic acid be represented by HX.

HXH⁺ +X^- [equilibrium]

ICE table

[HX] [H+] [X-]

initial 0.0884M 0 0

change -0.0884*0.25=-0.0221M +0.0221M +0.0221M

equilibrium 0.0884-0.0221=0.0663M 0.0221M 0.0221M

ka=equilibrium constant=[H+][X-]/[HX]=(0.0221M)(0.0221M)/(0.0663M)=0.00737M

ka=0.00737M

queen_honey_blossom answered 2 years ago

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