Question #9a945

1 Answer
anor277
Jun 21, 2017

We need pKa or Ka for acetic acid. And under other circumstances I might make you dig for it..........

Explanation:

This site tells me that pKa(HOAc)=4.76. Acetic acid is thus a fairly weak acid.......

And now we interrogate the equilibrium......

HOAc(aq)+H2O(l)⇌H3O++AcO−....

And thus we can write.......

[H3O+][AcO−][HOAc]=10−4.76

We know that, initially, [HOAc]=0.540⋅mol⋅L−1, and we guess that x⋅mol⋅L−1 of acetic acid undergoes protonolysis...

And thus 10−4.76=x×x0.540−x=x20.540−x

This is a quadratic in x, which we could solve exactly. Because chemists are lazy folk, however, we ASSUME that 0.540>>x, and thus 0.540−x≅0.540

And so x1=√10−4.76×0.540=0.00306

We can recycle this expression, i.e. we use successive approximations, to check on how good it was with respect to the initial assumptions.......

x2=√10−4.76×(0.540−0.00306)=0.00305

And this value has converged....so the approximation is good.

Now by definition of our problem x=[H3O+]=0.00305, and pH=2.52, and thus pOH=11.48.

And finally, [HO−]=10−11.48=3.27×10−12⋅mol⋅L−1.

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