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<pre style="white-space: pre-wrap; color: rgb(0, 0, 0); font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-thickness: initial; text-decoration-style: initial; text-decoration-color: initial;"><font face="Helvetica, Arial, sans-serif">Let's make sure that we get this right. It will really help if you look at the first page of the lab procedures when going through this.
For each of the <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>B</mi><mn>4</mn></msub><msub><mi>O</mi><mn>5</mn></msub><mo stretchy="false">(</mo><mi>O</mi><mi>H</mi><mo stretchy="false">)</mo><msup><sub><mn>4</mn></sub><mrow><mn>2</mn><mo>−</mo></mrow></msup></mrow></semantics></math> you need to use 2 H+. So the concentration of the <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>B</mi><mn>4</mn></msub><msub><mi>O</mi><mn>5</mn></msub><mo stretchy="false">(</mo><mi>O</mi><mi>H</mi><mo stretchy="false">)</mo><msup><mn>4</mn><mrow><mn>2</mn><mo>−</mo></mrow></msup></mrow></semantics></math> is going to be 1/2 the moles of acid used in the titration divided by the original volume you collected in the test tube (5 ml)
Now, the<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics> <annotation encoding="TeX"> </annotation></semantics></math>
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>K</mi><mrow><mi mathvariant="normal">s</mi><mi mathvariant="normal">p</mi></mrow></msub><mo>=</mo><mstyle mathvariant="normal"><mo stretchy="false">[</mo><mi mathvariant="normal">N</mi><msup><mi>a</mi><mo>+</mo></msup><msup><mo stretchy="false">]</mo><mn>2</mn></msup><mo stretchy="false">[</mo><msub><mi>B</mi><mn>4</mn></msub><msub><mi>O</mi><mn>5</mn></msub><mo stretchy="false">(</mo><mi mathvariant="normal">O</mi><mi mathvariant="normal">H</mi><msubsup><mo stretchy="false">)</mo><mn>4</mn><mrow><mn>2</mn><mo>−</mo></mrow></msubsup><mo stretchy="false">]</mo></mstyle><mo>=</mo><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><msup><mo stretchy="false">)</mo><mn>2</mn></msup><mi>x</mi></mrow><annotation encoding="TeX">K_{\mathrm{sp}}=\mathrm{[Na^+]^2 [B_4O_5(OH)_4^{2-}]}=(2x)^2 x</annotation></semantics></math>
The concentration of the <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>B</mi><mn>4</mn></msub><msub><mi>O</mi><mn>5</mn></msub><mo stretchy="false">(</mo><mi>O</mi><mi>H</mi><mo stretchy="false">)</mo><msup><mn>4</mn><mrow><mn>2</mn><mo>−</mo></mrow></msup></mrow></semantics></math> is going to be "x" and the concentration of the Na+ is going to be "2x",
and the value of <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mstyle mathvariant="normal"><msub><mi>K</mi><mrow><mi>s</mi><mi>p</mi></mrow></msub></mstyle><annotation encoding="TeX">\mathrm{K_{sp}}</annotation></semantics></math>is going to be "4x^3".
Got it?
When you prepare your plot of lnK vs 1/T, be sure to use KELVIN for the
temperature. Use 8.314 for R (with its correct units) when calculating
the enthalpy and entropy for the reaction.
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<div class="moz-signature">-- <br>
<b><em>Andrew J. Pounds, Ph.D.</em></b><br>
<em>Professor of Chemistry and Computer Science</em><br>
<em>Director of the Computational Science Program</em><br>
<em>Mercer University</em><br>
<em>1501 Mercer University Drive, Macon, GA 31207 </em><br>
<em>(478) 301-5627</em><br>
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