Important Concepts

1. Use The Three Step Method.
2. Simplify ratios before performing calculations with them. First, reduce ratios to lowest terms. For example, reduce 25:5 to 5:1. Then, convert ratio strengths to a percentage strength expressed as a decimal. For example, 1:5 turns into 0.2.
3. Use the formula: quantity x concentration = quantity x concentration
4. Concentration may be expressed in w/v as g/mL (100% = 1.0 g/mL = 100 g/100 mL), v/v as mL/mL (100% = 1.0 mL/mL = 100 mL/100 mL), or w/v as mg/mL (100 mg% = 1.0 mg/mL = 100 mg/100 mL). w/v stands for weight in volume concentration and v/v stands for volume in volume concentration.

Example

• Problem: If 50 mL of sodium chloride solution (0.5% w/v) is diluted to a liter, convert the resulting concentration into ppm.
• Known Variables; 50 mL of sodium chloride at 0.005 g/mL concentration, 1000 mL of diluted sodium chloride.
• Unknown Variables; X g/mL = concentration of 1000 mL diluted sodium chloride, Y ppm = concentration of 1000 mL diluted sodium chloride.
• Formula: quantity x concentration = quantity x concentration
• Solution:
  

  50 mL ( 0.005 ) = ( 1000 mL ) X 50 mL ( 0.005 ) = X 1000 mL X = 0.00025 0.00025 = 0.00025 = 0.00025 ( 100,000 ) = 25                                      1                1 ( 100,000 )    100,000          25 =              Y 100,000    1,000,000 25 x 1,000,000 = Y            100,000 Y = 250 ppm ppm = parts per million. Therefore, Y is parts per million where Y/1,000,000.

Alcohol
The volume of water and alcohol contracts when they are mixed. In other words, 10 mL of water mixed with 10 mL of alcohol will equal less than 20 mL of mixture. Therefore, it is not possible to predict the volume of water needed to dilute alcohol to a specific mL/mL strength. However, you can "qs" a solution to the proper volume.

• Problem: A pipette is graduated in units of 10 mL from 0 to 100 mL. How would you prepare the following?
  

  Rx	Rose Oil	2.5 mL
  	Alcohol qs	90 mL

• Known Variables; 2.5 mL of rose oil, 20 mL is the minimum amount measurable in a 100 mL graduated pipette. For reasons of accuracy, you should never measure an amount less than 20% of the capacity of the graduate.
• Explanation: 2.5 mL of rose oil cannot be measured directly as it is smaller than the smallest amount measurable by the graduate, 20 mL. Therefore, you have to measure 2.5 mL of rose oil indirectly by measuring 20 mL of a diluted solution of rose oil. Therefore, the strength of this dilution is 1:8 rose oil:alcohol as 2.5 mL x 8 = or > 20 mL.
• Unknown variable: X = the volume of the alcohol & rose oil mixture to dilute 20 mL of rose oil to.
• Solution:
  

  1 = 20 mL 8     X 8 ( 20 ) = X 1 X = 160 mL Measure 20 mL of rose oil. Dilute to 160 mL with alcohol. Then measure 20 mL of this diluted solution and qs to 90 mL with alcohol.

Acids
The strength of a concentrated acid is expressed in units of w/w. However, the strength of a diluted acid is expressed in units of w/v. Therefore, you have to include the specific gravity of a concentrated acid in your calculations when computing the volume of concentrated acid needed to compound a specific quantity of diluted acid.
Formula: ( mL dil acid ) ( % w/v ) = ( mL conc acid ) ( % w/w ) ( SG )

• Problem: You are requested to prepare 400 mL of phosphoric acid solution, 10% w/v. How many mL of a concentrated phosphoric acid solution (40% w/w, SG = 1.25) should be used?
• Known Variables; 400 mL dil phosphoric acid, 10% w/v dil phosphoric acid, 40% w/w conc phosphoric acid, SG = 1.25.
• Unknown Variable: X mL conc acid
• Formula: ( mL dil acid ) ( % w/v ) = ( mL conc acid ) ( % w/w ) ( SG )
• Solution:
  

  (400 mL)(0.1) = X mL (0.4)(1.25) (400 mL)(0.1) = X (0.4)(1.25) X = 80 mL

Milliequivalents
The milliequivalent (mEq) measures the concentration of chemical activity of a solution. It takes into account the valence of the ions in solution. 1 mEq = mg of solute expressed as 1/1000 of its gram equivalent weight.

• Problem: A patient is to receive 0.2 mEq of NaCl per kg of body weight. If the patient weighs 154 lb, how many mL of isotonic saline should be used (M.W. NaCl = 58.5) ?
• Known Variables; 154 lb patient, 0.2 mEq NaCl/kg body weight.
• Unknown Variable: X mL = isotonic saline used.
• Conversion Factors; 2.2 lb/kg, normal saline = 0.9 g/100 mL NaCl
• Solution:
  

  154 lb x 1 kg   x 0.2 mEq = 14.0 mEq               2.2 lb       1 kg NaCl molecular weight: 58.5 NaCl equivalent weight: 58.5 ÷ 1 = 58.5 1 is the valence of Na+ or Cl-. In other words, Na+ or Cl- has a +1 or -1 charge. Milliequivalent weight: 58.5 ÷ 1000 = 0.0585 g 0.0585 g =        X       1 mEq        14.0 mEq 0.0585 g x 14.0 mEq = X             1 mEq X = 0.819 g x 100 mL = 91.0 mL                          0.9 g

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