different specific gravities are mixed, the loss which one suffers will be balanced by the gain of the other; hence, the two liquids used must be mixed in inverse proportion to that existing between the gain and loss of specific gravity and the specific gravity of the mixture; the difference between the higher specific gravity and the desired specific gravity of the mixture will therefore indicate the proportion of the liquid having the lower specific gravity; and the difference between the lower specific gravity and the desired specific gravity will indicate the proportion of the liquid having the higher specific gravity. For example, if solution of ferric chloride, specific gravity 1.520, is to be reduced to 1.387 specific gravity by addition of a weaker solution of 1.280 specific gravity, 107 volumes of the stronger must be mixed with 133 volumes of the weaker solution, or, in other words, 1 volume of the former with 1.243 volumes of the latter. It is customary to set down a problem in alligation in the following manner to facilitate comparison: 1.387 1.520 0.107 = proportion of the stronger liquid. 1.280 0.133 proportion of the weaker liquid. If a definite volume of the mixture is desired, the requisite volume of the stronger and weaker liquids may be ascertained by dividing the desired volume by the sum of the proportionals, and then multiplying each proportional by the quotient so obtained; thus, if 32 fluidounces are wanted, divide 32 by 0.240 (0.107 +0.133), which yields 133.3; 0.107 × 133.3=14.27 fluidounces, the requisite volume of the stronger solution, and 0.133 x 133 3 = 17.73 fluidounces, the requisite volume of the weaker solution. To adjust the specific gravity of a given weight of a liquid to a higher or lower specific gravity, the following formula may be employed: in which a represents the weight of the diluent, w the weight of the liquid to be diluted, a the specific gravity of the liquid to be diluted, b the desired specific gravity, and e the specific gravity of the diluent. (Whenever water is the diluent, c is made 1.000.) As stated before weight of water equal in volume to w, weight of water equal in volume to æ, volume to w + x. To find the value of x, the following equation, b , Example: How much water must be added to 250 Gm. of solution of potassa of 1.539 specific gravity in order to reduce the specific gravity to 1.036? Substituting numerical values for the letters 250 x 1000/1 5291026) different specific gravities are mixed, the loss which one suffers will be balanced by the gain of the other; hence, the two liquids used must be mixed in inverse proportion to that existing between the gain and loss of specific gravity and the specific gravity of the mixture; the difference between the higher specific gravity and the desired specific gravity of the mixture will therefore indicate the proportion of the liquid having the lower specific gravity; and the difference between the lower specific gravity and the desired specific gravity will indicate the proportion of the liquid having the higher specific gravity. For example, if solution of ferric chloride, specific gravity 1.520, is to be reduced to 1.387 specific gravity by addition of a weaker solution of 1.280 specific gravity, 107 volumes of the stronger must be mixed with 133 volumes of the weaker solution, or, in other words, 1 volume of the former with 1.243 volumes of the latter. It is customary to set down a problem in alligation in the following manner to facilitate comparison : 1.387 { 1.520 0.107 proportion of the stronger liquid. 1.280, 0.133 proportion of the weaker liquid. If a definite volume of the mixture is desired, the requisite volume of the stronger and weaker liquids may be ascertained by dividing the desired volume by the sum of the proportionals, and then multiplying each proportional by the quotient so obtained; thus, if 32 fluidounces are wanted, divide 32 by 0.240 (0.107 +0.133), which yields 133.3; 0.107 x 133.3 14.27 fluidounces, the requisite volume of the stronger solution, and 0.133 × 133 3 = 17.73 fluidounces, the requisite volume of the weaker solution. To adjust the specific gravity of a given weight of a liquid to a higher or lower specific gravity, the following formula may be employed: we (a - b) x = a (b − e) in which a represents the weight of the diluent, w the weight of the liquid to be diluted, a the specific gravity of the liquid to be diluted, b the desired specific gravity, and e the specific gravity of the diluent. (Whenever water is the diluent, e is made 1.000.) As stated before (see page 63), weight of water equal in volume to w, = weight a of water equal in volume to x, volume to w + x. w + x x с weight of water equal in To find the value of x, the following equation, Example: How much water must be added to 250 Gm. of solution of potassa of 1.539 specific gravity in order to reduce the specific gravity to 1.036? Substituting numerical values for the letters 250 1.000 (1.539-1.036) 1.539 (1.036-1.000) in the above formula, we have x = ; To make a definite weight of a liquid of definite specific gravity by mixing two liquids of known specific gravity, both being of the same kind, or one being water: Let mw represent the desired weight of the mixture, the weight of the diluent, y the weight of the liquid to be diluted, and a, b, c the specific gravity of the liquid to be diluted, of the mixture desired, and of the diluent respectively. (Whenever water is the diluent, e is made 1.000.) Since x+y=mu, and the value of x the liquid to be diluted × c(ab) a (b−c) has been shown above to be the latter expression may be substituted for x in the equation, x + y = yx c(a-b) mw, thus +y= =mw. This simplified is yea― yeb + a (b-c) yab yac = mw × a (b—c), and cancelling, y × b (a —c) = mw × a (bc). y = mwa (bc) уса The value of y (weight of stronger liquid) having been ascertained, it is subtracted from mir, the desired weight of the mixture, to find the value of x, the weight of the diluent. Example: If it is desired to make 10 pounds of ammonia water of 0.960 specific gravity, from ammonia water of 0.900 specific gravity, mix 3.75 pounds of the latter with 6.25 pounds of water; for, substituting numerical values for the letters in the above formula, the weight of the liquid to be diluted is equal to For the adjustment of percentage in alcohol (by weight or volume), in acids (by weight), and in alkali solutions (by weight), the following rules may be applied: For reducing solutions from a higher to a lower percentage: Multiply the given quantity by the given percentage and divide by the required percentage; the quotient will be the quantity to which the liquid must be diluted by the addition of water. Since alcohol is frequently reduced in volume percentage, and contraction of volume invariably follows the admixture of alcohol and water, it becomes necessary, after contraction has ceased, to add sufficient water to restore the original volume of the mixture. Examples: Reduce 4 pints (64 fluidounces) of 93 per cent. (by volume) alcohol to 65 per cent.: 64 × 93 = 5952, and 5952 ÷ 65 = 91.57. Enough water must be added to the 4 pints of alcohol to yield, after contraction has ceased, 91.57 fluidounces. Reduce 2 pounds of hydrochloric acid from 31.9 per cent. to 10 per cent. 2 pounds = 32 avoirdupois ounces; 32 x 31.91020.8, and 1020.810=102.08. Enough water must be added to the 2 pounds of acid to bring the total weight up to 102.08 avoirdupois ounces. Reduce 8 troy ounces of stronger ammonia water, 28 per cent., to 10 per cent. strength: 8 X 28=224, and 22410-22.4. Enough water must be added to the 8 troy ounces of stronger ammonia water to bring the total weight up to 22.4 troy ounces. For making a definite quantity of a solution of a certain percentage by diluting a stronger solution with water: Multiply the required quantity by the required percentage, and divide by the higher percentage; the quotient will be the quantity of the stronger liquid necessary, and this subtracted from the total quantity required leaves the necessary quantity of water. When volume adjustment of alcohol is made, the same precautions in regard to contraction of volume must be observed as stated in the preceding rule. Examples: Make 1 gallon (128 fluid ounces) of 60 per cent. (by volume) alcohol from alcohol of 94 per cent. (by volume): 128 × 60 =7680, and 768094-81.7. Answer: 81.7 fluidounces of the stronger alcohol must be mixed with sufficient water to yield, after contraction has ceased, 128 fluidounces. Make 4 pounds of 25 per cent. phosphoric acid from the official 85 per cent. acid: 4 pounds = 64 av. ozs.; 64×25=1600, and 160085-18.823. Answer: Enough water must be mixed with 18.823 av. ozs. of the strong acid to bring the total weight up to 64 av. ozs. Make 720 grains of 5 per cent. caustic potash solution from a 12.5 per cent. solution: 720×5=3600, and 3600 ÷ 12.5288; 720288432. Answer: 288 grains of the 12.5 per cent. solution must be mixed with 432 grains of water. The adjustment of percentage in liquids may also be readily made by the process of alligation, as already explained under adjustment of specific gravities by volume, page 64. Pharmacists and drug jobbers are sometimes called upon to make mixtures of certain liquids or solids having different percentage strengths in order to produce a desirable average strength; this may be done by the general rule for alligation. Write the percentages in a column, and the mean percentage on the left. Connect the simples in pairs, one less than the mean with one greater; take the difference between the mean and the numbers representing the percentage strength of each simple and write it opposite the value with which it is linked. These differences are the relative quantities of the simples taken in the order in which their values stand. |