# Explanation of Chemical Calculations for Histology Preparation

## Making the Fixative: 2% glutaraldehyde in 0.1 M Na-cacodylate buffer with 0.1 M sucrose

The unit M is molarity, or mol/L.

### How much sodium cacodylate?

We want to create a 0.1 M buffer of sodium cacodylate. To calculate the amount of sodium cacodylate necessary, we need the molecular weight of sodium cacodylate, the final concentration of the solution, and the final volume.

The molecular weight of sodium cacodylate is 214.03 g/mol. The final concentration of the solution will be 0.1 M (0.1 mol/L) and the final volume will be 200 mL.

*mass sodium cacodylate* = (0.1 mol/L)*(214.03 g/mol)*0.2 L = 4.28 g

### How do I make 2% glutaraldehyde?

If you cannot purchase 2% glutaraldehyde, you will have to make some corrections.

For example, if we had 8% glutaraldehyde, we know that 8% is an indication of how many milliliters of glutaraldehyde to milliliters of water are in the solution; 8% is 0.08, or 8 mL out of 100 mL of solution.

We have the standard formula:

*Concentration _{1}**

*Volume*=

_{1}*Concentration**

_{2}*Volume*

_{2}

The volumes are up to our choosing and we know *Concentration _{1}* and

*Concentration*as 8% and 2%, respectively. However, we need to make numbers that we can actually measure; that is, not percentages.

_{2}We can say that there are 8 mL of glutaraldehyde for every 92 mL of water, creating an 8% solution. Furthermore, we know the density of glutaraldehyde to be 1.06 g/mL and that the molecular weight of glutaraldehyde is 100.12 g/mol. Using dimensional analysis, we can determine the concentration (as mol/L, or M) of the 8% solution:

*Concentration _{1}* = (8 mL/92 mL)*(1.06 g/mL)*(mol/100.12 g)*(1000 mL/L) = 0.9206 M

Similarly, we can determine the concentration of the 2% solution, *Concentration _{2}*:

*Concentration _{2}* = (2 mL/98 mL)*(1.06 g/mL)*(mol/100.12 g)*(1000 mL/L) = 0.2161 M

How much of the 8% solution do we need? We can arbitrarily set our desired final volume to 100 mL, and rearrange our initial equation to:

*Volume _{1}* =

*Concentration**

_{2}*Volume*/

_{2}*Concentration*

_{1}

Substituting in our values:

*Volume _{1}* = (0.2161 M)*(100 mL)/0.9206 M = 23.5 mL

So, to create 100 mL of a 2% glutaraldehyde solution from an 8% solution, add 23.5 mL of 8% glutaraldehyde to 76.5 mL of the prepared Na-cacodylate buffer.

### How much sucrose?

The steps to calculating the amount of sucrose needed to make a 0.1 M solution are the same for calculating the amount of sodium cacodylate. The molecular weight of sucrose is 342.3 g/mol, the final concentration is 0.1 M, and the final volume is 100 mL.

*mass sucrose* = (0.1 mol/L)*(342.3 g/mol)*0.1 L = 3.42 g

## Making the Treatment: 1% tannic acid in 0.15 M Na-cacodylate buffer

### How do I make a 1% solution of tannic acid?

To make a 1% solution of tannic acid, simply add 1 g of tannic acid to 100 mL of dH_{2}O. The ratio is not exact, but it is close enough for this purpose.

### How do I make a 0.15 M sodium cacodylate buffer?

We know the final volume of the solution will be 100 mL, so we just need to figure out how many grams of sodium cacodylate will need to be added to the dH_{2}O. To do this, we can use the molecular weight of sodium cacodylate: 214.03 g/mol.

*mass sodium cacodylate* = (0.15 mol/L)*(214.03 g/mol)*(1 L/1000 mL)*100 mL = 3.21 g

Therefore, to make the treatment, you will add 1 g of tannic acid and 3.21 g of sodium cacodylate to 100 mL of dH_{2}O.

## Making the Post-fixative: 1% osmium tetroxide in 0.1 M Na-cacodylate buffer

In making the fixative, we made an excess of 0.1 M sodium cacodylate buffer, of which at least 100 mL should have been set aside. Osmium tetroxide is a very hazardous chemical, so we want to work with as little of it as possible. If you cannot purchase 1% osmium tetroxide, you will need to do some conversions.

We were not able to find 1% osmium tetroxide, so we used to 2% instead leaving us with the question:

### How much 2% osmium tetroxide makes a 1% solution?

Resorting back to our favorite equation:

*Concentration _{1}**

*Volume*=

_{1}*Concentration**

_{2}*Volume*

_{2}

We know *Concentration _{1}* to be 2%, or 0.02, and

*Concentration*to be 1%, or 0.01. We can arbitrarily choose a

_{2}*Volume*of 25 mL. Working out the math, we find that

_{2}*Volume*, the volume of 2% osmium tetroxide, should be 12.5 mL. Logically, this makes sense because we are cutting the concentration in half, the volume of the 2% osmium tetroxide should be one half of the final solution volume.

_{1}So, to make the post-fixative, combine equal amounts of osmium tetroxide and 0.1 M sodium cacodylate. For our case, we used 12.5 mL of each.

## Making the Graded Ethanol Series

To dry the samples, the tissues are moved to solutions of increasing ethanol concentration.

We only wanted to make 40 ml solutions, because that is more than enough for the tiny samples we made, so 95% (190 proof) ethanol must be diluted to make lesser concentrations. So, how many mL of 95% ethanol does it take to make a 40 mL solution of 30% ethanol? We can use the formula:

*Concentration _{1}**

*Volume*=

_{1}*Concentration**

_{2}*Volume*

_{2}

Where:

*Concentration _{1}* = 95% = 0.95

*Concentration*= 30% = 0.30

_{2}*Volume*= the amount of 95% ethanol necessary to make a 30% solution

_{1}*Volume*= 40 mL

_{2}To solve for *Volume _{1}*:

*Volume _{1}* =

*Concentration**

_{2}*Volume*/

_{2}*Concentration*

_{1}= 0.30*40 mL/0.95 = 12.6 mL ethanol

To create the 30% solution, we will add 12.6 mL of ethanol to 27.4 mL of dH_{2}O (40-12.6 mL). The rest of the series were calculated similarly:

40% = 16.8 ml of 95% EtOH

50% = 21.05 ml of 95% EtOH

70% = 29.5 ml of 95% EtOH

80% = 33.7 ml of 95% EtOH

90% = 37.9 ml of 95% EtOH

For the 100% wash, use anhydrous 200 proof ethanol. Once the bottle has been unsealed, the ethanol is no longer 100%.