pka log but we don t have pka. We have Ka. To determine pka So, now we have everything we need:
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1 What is the of the following solution: 0.3M boric acid (H) and 0.9M sodium borate ( - )? Ka = 6.4x10-10 pka log H but we don t have pka. We have Ka. To determine pka pka log( Ka) log( ) 9.19 pka 9.19 H So, now we have everything we need: pka log H log What is the of a solution containing 0.2M weak basic drug and 0.002M of its salt. The pka=9.36 at 25 o C. We know that - is the base and H is the acid. Here, - is the weak basic drug and H is its (conjugate = acidic) salt. pka H So, we just plug and chug. PH pka log H 9.36 log
2 What is the of a basic drug and its conjugate salt, if their concentrations are 0.2M and 0.4M, respectively? ssume pka=8.46. pka H 0.4 PH pka log H log What is the of a solution containing 0.3M ammonium hydroxide and 0.015M of ammonium chloride? Kb=1.8x10-5 at 25 o C. Kw= So who s my base and who s my acid? et s write out each guy to find out. mmonium hydroxide = NH 4 OH mmonium chloride = NH 4 Cl ny guy with OH is always a base (like Ca(OH) 2 or NaOH, etc). so, if NH 4 OH is the base, that makes NH 4 Cl its conjugate acidic salt. I m going to stick with the - /H notation just because I m used to it, but feel free to use Dr. tef s base and salt notation. It s the same thing. 0.3 Kb So far we know: H pka log H but we need pka for H-H, not Kb, so we need to covert it. pkb First switch Kb to pkb log( Kb) log( ) 4.74 To get pka, we use our knowledge that pkw 14 pka pkb
3 and with a little switcheroo pka 14 pkb Now we have everything we need: 0.3 H pka log a) What is the molar ratio of salt/acid required to adjust the of a solution to 5.5 using Na acetate and acetic acid solution. Ka of acetic acid = 1.8x10-5 at 25 o C. b) How much Na acetate (MW=82) and acetic acid (MW=60) are required to adjust the of a solution to 5.5 if the total buffer molar concentration is 3.35x10-2 mol/. (see below) a) We re looking for the molar ratio of salt/acid. If we have an acid, its conjugate salt will be basic, so we re solving for the ratio of - /H. So far, we have 5.5 Ka pka log H pka First we ll determine pka from Ka: log( Ka ) Rearranging the equation for - /H: 4.75
4 5.5 pka 4.75 pka log H pka H :1 If you don t round, I think you get an answer like If the prof asks us for the ratio strength of each, first we can determine the percentage of each guy and then go from there H 1 total % 100% 85.2% % H 100% 14.8% 6.75 To find the ratio strength, I just push the inverse button on my calc. % 85.2% ratio 1/ ratio strength % H 14.8% ratio 1/ H ratio strength We already knew the second guy from above.
5 b. How much Na acetate (MW=82) and acetic acid (MW=60) are required to adjust the of a solution to 5.5 if the total buffer molar concentration is 3.35x10-2 mol/. So we already know the percentages of each guy, so (1) we ll just apply that to the molarity that we re given here and (2) convert to grams after that. 85.2% ( ) of 14.8% ( H) of mol mol mol mol H Remember that Na acetate is the base ( - ) and acetic acid is the acid (H). Refer to molecular weights above mol mol 82g 1mol 2.34 g H 60g H mol H g 2.34 g H Na acetate g acetic acid 300m of 1% atropine sulfate is prepared. How many grams of sodium chloride should be added to make the solution isotonic? The sodium chloride equivalent of atropine is First determine the number of grams of atropine you have knowing the concentration is 1% (1g/100m). 300m 1g atropine 100m 3g atropine Next determine the number of grams that the atropine is equivalent to. 3g atropine 0.13g 1g atropine 0.39g Now determine the number of grams of that would be required for the entire solution (300m) to be isotonic (0.9% ). 300m 0.9g 100m 2.7g If we need a total of 2.7g and we only have the equivalent of 0.39g, then we need to add:
6 2.7g 0.39g 2. 31g additional Isotonic sodium chloride solution contains 0.9%. If a sodium chloride equivalent of boric acid is 0.52, what is the percentage strength of an isotonic solution of boric acid? If a solution is 0.9%, then there are 0.9g per 100m of solution. 0.9g 100m 1g B 0.52g 1.73g B 100m 1.73% B If the sodium chloride equivalent of Drug is 0.1, is an 8% solution of Drug hypo-, hyper-, or iso-tonic? n 8% solution of Drug is composed of 8g/100m. 0.8g Drug 100m 0.1g 1g Drug 0.8g 100m 0.8% solution is considered isotonic at 0.9%. Because 0.8% is BEOW 0.9%, the solution is said to be HYPOtonic. Naphazoline HCl 1% qs Water ad 30m Dispense isotonic solution. The sodium chloride equivalent of Naphazoline is If a 30m solution is 1% naphazoline, then there are 3g Naphazoline: 30m 1g Naphazoline 100m 0.3g Naphazoline Determine how many grams this is equivalent to.
7 0.3g Naphazoline 0.27g 1g Naphazoline 0.081g This is how many equivalent grams of we have. If we want the solution to be isotonic (0.9% ) then we need 0.27g. 30m 0.9g 100m 0.27g So how many grams of did we need to add? We have 0.081g and we need 0.27g. 0.27g 0.081g g equivalent Phenacaine HCl 1% 0.2 Chlorbutanol 0.5% 0.14 Boric acid qs 0.52 Purified water qs 60m Prepare isotonic solution. We re going to determine the equivalent number of grams of of each ingredient in the 60m solution. 60m 1g Phenacaine 100m 0.2g 1g Phenacaine 0.12g 60m 0.5g Chlorbuta nol 100m 0.14g 1g Chlorbuta nol Determine the total number of equivalent g 0.12g 0.042g g Determine the total number of grams of required for an isotonic (0.9% ) solution. 60m 0.9g 100ml 0.54g
8 So we need 0.54g for an isotonic solution, but we only have 0.162g. Determine the number of grams of additional to be added. 0.54g 0.162g g We need an equivalent of 0.378g, but we re qs ing with boric acid ( equivalent = 0.52) g 1g B 0.52g 0.727g B How many mg of are required to prepare the following prescription of isotonic epinephrine solution? The epinephrine solution is isotonic (has an 0.9% equivalent). tropine sulfate (E=0.14) 1% Epinephrine solution 1:2000 8m Purified water qs 30m Determine the number of grams of equivalent we have in the two ingredients. 30m 1g tropine 100m 0.14g 1g tropine 0.042g 30m 0.9g 100m 0.072g Determine the total number of equivalent g 0.072g g Determine the total number of grams of required for an isotonic (0.9% ) solution. 30m 0.9g 100ml 0.27g So we need 0.27g for an isotonic solution, but we only have 0.114g. Determine the number of grams of additional to be added. 0.27g 0.114g g
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