User:Matt Hartings/Notebook/AU Biomaterials Design Lab/2016/08/29: Difference between revisions
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|style="background-color: #EEE"|[[Image:BDLlogo_notext_lr.png|128px]]<span style="font-size:22px;"> Biomaterials Design Lab</span> | |style="background-color: #EEE"|[[Image:BDLlogo_notext_lr.png|128px]]<span style="font-size:22px;"> Biomaterials Design Lab</span> | ||
|style="background-color: #F2F2F2" align="center"| | |style="background-color: #F2F2F2" align="center"|[[File:Report.png|frameless|link={{#sub:{{FULLPAGENAME}}|0|-11}}]][[{{#sub:{{FULLPAGENAME}}|0|-11}}|Main project page]]<br />{{#if:{{#lnpreventry:{{FULLPAGENAME}}}}|[[File:Resultset_previous.png|frameless|link={{#lnpreventry:{{FULLPAGENAME}}}}]][[{{#lnpreventry:{{FULLPAGENAME}}}}{{!}}Previous entry]] }}{{#if:{{#lnnextentry:{{FULLPAGENAME}}}}|[[{{#lnnextentry:{{FULLPAGENAME}}}}{{!}}Next entry]][[File:Resultset_next.png|frameless|link={{#lnnextentry:{{FULLPAGENAME}}}}]]}} | ||
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==Description== | ==Description== | ||
<u>stock solutions</u> | <u>stock solutions</u> | ||
# 1 L of | # 1 L of 0.001 M HCl | ||
## Concentrated HCl should be: 11.65 M | ## Concentrated HCl should be: 11.65 M | ||
## M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | ## M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | ||
## (11.65 M)(V<sub>1</sub>) = ( | ## (11.65 M)(V<sub>1</sub>) = (0.001M)(1L) | ||
## V<sub>1</sub> = 85.8 | ## V<sub>1</sub> = 85.8 μL | ||
# 1 L of | # 1 L of 0.001 M HNO<sub>3</sub> | ||
## Concentrated HNO<sub>3</sub> should be: 15.8 M | ## Concentrated HNO<sub>3</sub> should be: 15.8 M | ||
## M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | ## M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | ||
## (15.8 M)(V<sub>1</sub>) = ( | ## (15.8 M)(V<sub>1</sub>) = (0.001M)(1L) | ||
## V<sub>1</sub> = 63.2 | ## V<sub>1</sub> = 63.2 μL | ||
# 1 L of 1 M NaOH | # 1 L of 1 M NaOH | ||
## MW of NaOH = 39.997 g/mol | ## MW of NaOH = 39.997 g/mol | ||
Line 30: | Line 30: | ||
## 1 M = (m/39.997 g/mol)/1L | ## 1 M = (m/39.997 g/mol)/1L | ||
## m = 39.997 g | ## m = 39.997 g | ||
# 1 L of 0.001 M NaOH | |||
## Use 1M stock solution | |||
## M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
## (1 M)(V<sub>1</sub>) = (0.001M)(1L) | |||
## V<sub>1</sub> = 1 mL | |||
# 10 mL of 12.5 mM Fructose | # 10 mL of 12.5 mM Fructose | ||
## MW = 180.16 g/mol | ## MW = 180.16 g/mol | ||
Line 38: | Line 43: | ||
## 0.0025 M = (m/303.33 g/mol)/(0.010 L) | ## 0.0025 M = (m/303.33 g/mol)/(0.010 L) | ||
## m = 7.58 mg | ## m = 7.58 mg | ||
# Protein (BSA) | |||
## MW = 66776 g/mol | |||
## 30 μM = (m/66776 g/mol)/(0.010 L) | |||
## m = 20.0 mg | |||
<u> sample solutions </u> | <u> sample solutions </u> | ||
Line 45: | Line 54: | ||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | ### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | ||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | ### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | ||
## Volume of fructose stock | |||
### V<sub>fructose stock</sub> = 0 mL | |||
## Volume of HCl stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-4</sup> M)(5 mL) | |||
### V<sub>HCl stock</sub> = 0.5 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>HCl stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.483 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 5 ([H<sup>+</sup>] = 1x10<sup>-5</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### V<sub>fructose stock</sub> = 0 mL | |||
## Volume of HCl stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-5</sup> M)(5 mL) | |||
### V<sub>HCl stock</sub> = 0.05 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>HCl stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.933 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 6 ([H<sup>+</sup>] = 1x10<sup>-6</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### V<sub>fructose stock</sub> = 0 mL | |||
## Volume of HCl stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-6</sup> M)(5 mL) | |||
### V<sub>HCl stock</sub> = 0.005 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>HCl stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.978 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 7 ([H<sup>+</sup>] = 1x10<sup>-7</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### V<sub>fructose stock</sub> = 0 mL | |||
## Volume of HCl stock | |||
### V<sub>HCl stock</sub> = 0 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>HCl stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.983 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 8 ([H<sup>+</sup>] = 1x10<sup>-8</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### V<sub>fructose stock</sub> = 0 mL | |||
## Volume of 0.001 M NaOH stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-6</sup> M)(5 mL) | |||
### V<sub>NaOH stock</sub> = 0.005 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>NaOH stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.978 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 9 ([H<sup>+</sup>] = 1x10<sup>-9</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### V<sub>fructose stock</sub> = 0 mL | |||
## Volume of 0.001 M NaOH stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-5</sup> M)(5 mL) | |||
### V<sub>NaOH stock</sub> = 0.05 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>NaOH stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.933 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 10 ([H<sup>+</sup>] = 1x10<sup>-10</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### V<sub>fructose stock</sub> = 0 mL | |||
## Volume of 0.001 M NaOH stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-4</sup> M)(5 mL) | |||
### V<sub>NaOH stock</sub> = 0.5 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>NaOH stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.483 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 11 ([H<sup>+</sup>] = 1x10<sup>-11</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### V<sub>fructose stock</sub> = 0 mL | |||
## Volume of 1 M NaOH stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1 M)V<sub>1</sub> = (1x10<sup>-3</sup> M)(5 mL) | |||
### V<sub>NaOH stock</sub> = 0.005 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>NaOH stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.978 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 12 ([H<sup>+</sup>] = 1x10<sup>-12</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### V<sub>fructose stock</sub> = 0 mL | |||
## Volume of 1 M NaOH stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1 M)V<sub>1</sub> = (1x10<sup>-2</sup> M)(5 mL) | |||
### V<sub>NaOH stock</sub> = 0.05 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>NaOH stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.933 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 4 ([H<sup>+</sup>] = 1x10<sup>-4</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0125 M)V<sub>1</sub> = (0.0000625 M)(5 mL) | |||
### V<sub>fructose stock</sub> = 0.025 mL | |||
## Volume of HCl stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-4</sup> M)(5 mL) | |||
### V<sub>HCl stock</sub> = 0.5 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>HCl stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.458 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 5 ([H<sup>+</sup>] = 1x10<sup>-5</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0125 M)V<sub>1</sub> = (0.0000625 M)(5 mL) | |||
### V<sub>fructose stock</sub> = 0.025 mL | |||
## Volume of HCl stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-5</sup> M)(5 mL) | |||
### V<sub>HCl stock</sub> = 0.05 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>HCl stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.908 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 6 ([H<sup>+</sup>] = 1x10<sup>-6</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0125 M)V<sub>1</sub> = (0.0000625 M)(5 mL) | |||
### V<sub>fructose stock</sub> = 0.025 mL | |||
## Volume of HCl stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-6</sup> M)(5 mL) | |||
### V<sub>HCl stock</sub> = 0.005 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>HCl stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.953 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 7 ([H<sup>+</sup>] = 1x10<sup>-7</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0125 M)V<sub>1</sub> = (0.0000625 M)(5 mL) | |||
### V<sub>fructose stock</sub> = 0.025 mL | |||
## Volume of HCl stock | |||
### V<sub>HCl stock</sub> = 0 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>HCl stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.958 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 8 ([H<sup>+</sup>] = 1x10<sup>-8</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0125 M)V<sub>1</sub> = (0.0000625 M)(5 mL) | |||
### V<sub>fructose stock</sub> = 0.025 mL | |||
## Volume of 0.001 M NaOH stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-6</sup> M)(5 mL) | |||
### V<sub>NaOH stock</sub> = 0.005 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>NaOH stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.953 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 9 ([H<sup>+</sup>] = 1x10<sup>-9</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0125 M)V<sub>1</sub> = (0.0000625 M)(5 mL) | |||
### V<sub>fructose stock</sub> = 0.025 mL | |||
## Volume of 0.001 M NaOH stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-5</sup> M)(5 mL) | |||
### V<sub>NaOH stock</sub> = 0.05 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>NaOH stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.908 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 10 ([H<sup>+</sup>] = 1x10<sup>-10</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0125 M)V<sub>1</sub> = (0.0000625 M)(5 mL) | |||
### V<sub>fructose stock</sub> = 0.025 mL | |||
## Volume of 0.001 M NaOH stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1x10<sup>-3</sup> M)V<sub>1</sub> = (1x10<sup>-4</sup> M)(5 mL) | |||
### V<sub>NaOH stock</sub> = 0.5 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>NaOH stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.458 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 11 ([H<sup>+</sup>] = 1x10<sup>-11</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0125 M)V<sub>1</sub> = (0.0000625 M)(5 mL) | |||
### V<sub>fructose stock</sub> = 0.025 mL | |||
## Volume of 1 M NaOH stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1 M)V<sub>1</sub> = (1x10<sup>-3</sup> M)(5 mL) | |||
### V<sub>NaOH stock</sub> = 0.005 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>NaOH stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.953 mL | |||
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 12 ([H<sup>+</sup>] = 1x10<sup>-12</sup> M), 5 mL total | |||
## Volume of AuCl<sub>3</sub> stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL) | |||
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL | |||
## Volume of fructose stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (0.0125 M)V<sub>1</sub> = (0.0000625 M)(5 mL) | |||
### V<sub>fructose stock</sub> = 0.025 mL | |||
## Volume of 1 M NaOH stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (1 M)V<sub>1</sub> = (1x10<sup>-2</sup> M)(5 mL) | |||
### V<sub>NaOH stock</sub> = 0.05 mL | |||
## Volume of Protein stock | |||
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> | |||
### (30 μM)V<sub>Protein Stock</sub> = (3.1 μM))(5 mL) | |||
### V<sub>Protein Stock</sub> = 0.517 mL | |||
## Volume of water | |||
### V<sub>water</sub> = 5 mL - V<sub>AuCl<sub>3</sub> stock</sub> - V<sub>fructose stock</sub> - V<sub>NaOH stock</sub> - V<sub>Protein Stock</sub> | |||
### V<sub>water</sub> = 3.908 mL | |||
==Data== | ==Data== |
Latest revision as of 01:52, 27 September 2017
Biomaterials Design Lab | Main project page Previous entry Next entry |
ObjectiveCalculations for stock solutions to be made tomorrow. Descriptionstock solutions
sample solutions
Data
NotesThis area is for any observations or conclusions that you would like to note.
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