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<p><font face="serif">So I have gotten several questions today about
4.19. Many of you are confusing K with Kw.</font></p>
<p><font face="serif"><math
xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>K</mi><mi>w</mi></msub><mo>=</mo><mo
stretchy="false">[</mo><msub><mi>H</mi><mn></mn></msub><msup><mi></mi><mo>+</mo></msup><mo
stretchy="false">]</mo><mo stretchy="false">[</mo><mi>O</mi><msup><mi>H</mi><mo>−</mo></msup><mo
stretchy="false">]</mo></mrow><annotation encoding="TeX">K_w=[H_3O^+][OH^-]</annotation></semantics></math><br>
</font></p>
<p><font face="serif">You need to find K for the dissociation of
water, which has to include the molarity of water.</font></p>
<p><font face="serif"><math
xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>K</mi><mo>=</mo><mfrac><mrow><mo
stretchy="false">[</mo><msup><mi>H</mi><mo>+</mo></msup><mo
stretchy="false">[</mo><mi>O</mi><msup><mi>H</mi><mo>−</mo></msup><mo
stretchy="false">]</mo></mrow><mrow><mo
stretchy="false">[</mo><msub><mi>H</mi><mn>2</mn></msub><mi>O</mi><mo
stretchy="false">]</mo></mrow></mfrac></mrow><annotation
encoding="TeX">K=\frac{[H^+[OH^-]}{[H_2O]}</annotation></semantics></math><br>
</font></p>
<div class="moz-signature">The molarity of H+ and OH- you can find
from the pH. The molarity of water is found by dividing the moles
of water by volume. 1 L of water weighs 1000 g. 1000 g / 18 g/mol
= 55.6 molar.</div>
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<div class="moz-signature">For the last part you can look up the
enthalpy of the reactin in table 4.2 and then use equation 4.53 in
your text to compute the equilibrium constant at the other
temperature.</div>
<div class="moz-signature"><br>
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<div class="moz-signature"><br>
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<div class="moz-signature">-- <br>
<b><i>Andrew J. Pounds, Ph.D.</i></b><br>
<i>Professor of Chemistry and Computer Science</i><br>
<i>Director of the Computational Science Program</i><br>
<i>Mercer University, Macon, GA 31207 (478) 301-5627</i></div>
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