peÅna wersja - pdf - Polimery w Medycynie
peÅna wersja - pdf - Polimery w Medycynie
peÅna wersja - pdf - Polimery w Medycynie
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<strong>Polimery</strong>51pH-meterSpH-meterRESULTS AND DISCUSSIONFedorov and Shmata [17] evaluated followingequation for resolving the pH of a solution of salt ofweak base and strong acid, titrated by an acid:DARyc. 1. Schemat systemu zastosowanego do badaniazmian pH, wraz ze zmianami temperatury, w preparaciez chlorowodorkiem lidokainy: D – kompartmentdonorowy, A – kompartment akceptorowy, S –błona półprzepuszczalnaFig. 1. Scheme of device applied for the assessementof pH changes in the preparation of lidocaine hydrochloride:D – donor compartment, A – acceptor compartment,S – semipermeable membraneazotu. pH mierzono na początku badania oraz po 24godzinach, w podanych wyżej warunkach temperatury.Schemat systemu przedstawiono na rycinie 1.Obrazowaniew mikroskopie optycznymPróbki poddano badaniu w mikroskopie optycznymOlympus BX51, z okularami 20 × 0.46 mm i 50× 0.80 mm, a następnie obraz nagrywano w kamerzecyfrowej Olympus DP12, U-TVO.5XC-2, w świetledziennym.WYNIKI I DYSKUSJAFedorov i Shmata [17] zaproponowali do ocenyodczynu środowiska – pH – następujące równanie:pH = pK w + {ln[fna(x)] – (v 1 +x)}/ln10 (1)K w przedstawia iloczyn jonowy wody, x to zmiennaopisująca objętość titranta dodanego w odpowiednimpunkcie miareczkowania, a wartość fna(x) = xn 2--v 1 n 1 , gdzie v 1 jest objętością roztworu analitu, n 1 jestnormalnością roztworu analitu, n 2 jest normalnościąroztworu titranta. Przedstawione zależności zostałypH = pK w + {ln[fna(x)] – (v 1 + x)}/ln10 Eq. (1)The K w represents the ionic product of water,and x represents the variable describing the volumeof the titrant added at a given titration point. Thevalue fna(x) = xn 2 –v 1 n 1 ,where v 1 is the volume of theanalyte solution, n 1 is the normality of the analytesolution, n 2 is the normality of titrant solution. Theequation was adapted to perform respective calculationsfor pH in systems comprising of anionic polymerand lidocaine.The predicted – due to the calculations – pH oflidocaine solution, in the presence of hydrochloricacid and polyacrylic acid is presented on the Figure2 – left part. In the assumed conditions of experimentthe pH of the solution in equimolar compositionfor HCl is ca. 5,03 reaching 2,30 after doubleamount of HCl. When polyacrylic acid is applied inthe salt composition, the estimated value of pH inequimolar ratio is ca. 6,95 as it is usually for the saltsof weak acid and weak base. The pH of mixture withtwo-fold amount of polyacrylic acid is approaching6,0 pH. The estimated values of pH in acceptor compartment,when respective dissolution is considered,are presented on the Fig – right part. The pH value inthe case of dissolution of LD-PA should increasefrom ca. 4,25 to around 4,76 whereas in the case ofLD-HCl it should rise from 4,95 to 6,20.The prepared formulations, containing lidocainehydrochloride, were heated, and pH of the formulationswas recorded at increasing three temperatures.The formulations were assessed before and after dialysisperformed through a semipermeable membrane.The initial values of pH were assigned as “ini”,and the final values of pH after 24 hrs of dialysis weredescribed as “fin” on the Figure 3. In the case of a systemcontaining lidocaine and polyacrylic acid the pHin donor compartment increased slightly, in thecourse of release of the drug – Figure 3, and the differencebetween pH at the beginning of the processand at the final stage was ca. 0,75 pH unit. Oppositely,in the methylcellulose and lidocaine formulationsthe pH increased ca. 2 units, and acquired the6,5 pH level. The pH values increased only slightlywith the temperature increase.The pH observed in donor compartment duringthe assessments increased, when the drug was released