13.1 Metric System, Percent Solutions & Dilutions

13.1 Metric System, Percent Solutions & Dilutions

ASCP HT Exam Focus: The ASCP BOC content outline lists Laboratory Mathematics with three sub-topics: metric system, percent solutions/dilutions, and molar solutions. Calculation items are scored like any other question, and they are the most reliably earnable points on the examination because the answer is deterministic.

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The Metric System

Conversions used daily in histology

• $1\text{ L} = 1000\text{ mL}$; $1\text{ mL} = 1000\,\mu\text{L}$
• $1\text{ g} = 1000\text{ mg}$; $1\text{ mg} = 1000\,\mu\text{g}$
• $1\text{ mm} = 1000\,\mu\text{m}$, so a 4 micrometre section is 0.004 mm, and a 5 mm grossed slice is 5000 micrometres
• $1\text{ ppm} = 1\text{ mg/L} = 1\,\mu\text{g/mL}$ - the unit used for formaldehyde and xylene air monitoring
• Temperature: $^{\circ}\text{C} = (^{\circ}\text{F} - 32) \times 5/9$. A 60 degrees C paraffin oven is 140 degrees F.

Exam trap: cassette and mold dimensions are given in millimetres, microtome settings in micrometres. A specimen grossed at 3 mm is 750 times thicker than the 4 micrometre section cut from it, which is exactly why penetration and processing schedules are thickness-driven.

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Percent Solutions: Three Conventions

The w/v formula

$$\text{Grams} = \frac{\%\ \times\ \text{final volume in mL}}{100}$$

Worked example. Prepare 800 mL of 2.5 percent (w/v) potassium metabisulfite for a silver method: $\text{Grams} = (2.5 \times 800)/100 = \mathbf{20.0\text{ g}}$, dissolved in water and brought to 800 mL.

The v/v formula

$$\text{mL solute} = \frac{\%\ \times\ \text{final volume in mL}}{100}$$

Worked example. Prepare 500 mL of 1 percent (v/v) acid alcohol from concentrated hydrochloric acid and 70 percent ethanol: $(1 \times 500)/100 = \mathbf{5.0\text{ mL}}$ of concentrated HCl added slowly to 495 mL of 70 percent ethanol. Always add the acid to the alcohol, never the reverse.

"Bring to volume" versus "add to": a true percent solution is made by dissolving the solute in part of the solvent and then bringing the total to the final volume in a graduated cylinder or volumetric flask. Adding solute to a full final volume of solvent overshoots the volume and under-concentrates the reagent. For dilute working solutions the difference is small; for concentrated silver and dye solutions it is not.

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The Formalin Trap

10 percent formalin is not 10 percent formaldehyde.

• Commercial "formalin" is a saturated aqueous solution containing 37 to 40 percent formaldehyde gas with methanol as a stabilizer.
• 10 percent formalin means 100 mL of that commercial stock diluted to 1000 mL, which delivers 3.7 to 4.0 percent actual formaldehyde.
• 10 percent neutral buffered formalin (NBF) is the same 1:10 dilution made up in phosphate buffer to pH 6.8 to 7.2.

This distinction is tested directly, and it also explains why a fixative labelled "4 percent paraformaldehyde" and one labelled "10 percent formalin" are chemically equivalent in formaldehyde content.

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Ratio and Serial Dilutions

A ratio dilution written 1:10 means one part of the substance in a total of ten parts, which is one part solute plus nine parts diluent. The dilution factor is 10, so the concentration falls to one tenth.

Worked example - antibody titre. An immunohistochemistry protocol calls for a 1:400 working dilution in a final volume of 12 mL:

$$V_{\text{stock}} = \frac{12\text{ mL}}{400} = 0.03\text{ mL} = \mathbf{30\,\mu\text{L}}$$

Pipette 30 microlitres of antibody concentrate into 11.97 mL of antibody diluent.

Serial dilution. Antibody titration is performed as a doubling or tenfold series: 1:100, 1:200, 1:400, 1:800. Each tube carries forward a fixed volume into an equal or ninefold volume of diluent, so the total dilution is the product of the individual steps (a 1:10 followed by a 1:20 is a 1:200).

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The Dilution Equation

$$C_1V_1 = C_2V_2 \quad\Rightarrow\quad V_1 = \frac{C_2 \times V_2}{C_1}$$

where $C_1$ is the stock concentration, $V_1$ the stock volume required, $C_2$ the desired concentration, and $V_2$ the desired final volume. The diluent volume is $V_2 - V_1$.

Worked example - graded alcohols. Prepare 1000 mL of 70 percent ethanol from 95 percent stock:

$$V_1 = \frac{70 \times 1000}{95} = 736.8\text{ mL of 95 percent stock} + 263.2\text{ mL water}$$

Worked example - from absolute alcohol. Prepare 500 mL of 80 percent ethanol from 100 percent stock:

$$V_1 = \frac{80 \times 500}{100} = 400\text{ mL absolute} + 100\text{ mL water}$$

Worked example - silver stock. Prepare 250 mL of 2 percent silver nitrate from a 20 percent stock:

$$V_1 = \frac{2 \times 250}{20} = 25\text{ mL stock} + 225\text{ mL deionized water}$$

Unit discipline: $C_1$ and $C_2$ must share one unit, and $V_1$ and $V_2$ must share another. Mixing percent with molar, or millilitres with litres, is the single most common calculation error on the examination.