Application of method for determining cloud point and free sugar buffer coefficient of iron- carbohydrate complex
A technology of carbohydrates and buffer coefficients, applied in the direction of testing pharmaceutical preparations, material inspection products, etc., can solve the problems of operator conditions and test environment influence, turbidity analysis equipment test value error, etc.
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Embodiment 1
[0083] Example 1 Example of Determination Method for Iron Sucrose Cloud Point (PP) and Free Sugar Coefficient (n)
[0084] Preparation of iron sucrose test solution: Take 1 g of iron sucrose injection, put it in a 300ml beaker, add 200ml of water to dilute. Preparation of acid solution: Take concentrated hydrochloric acid and add purified water to prepare 0.1mol / L hydrochloric acid solution. Under the condition of continuous stirring, add 0.1 mol / L hydrochloric acid solution dropwise to the iron sucrose test solution. At different pH, the turbidimeter measures the turbidity of the solution, and records the pH and turbidity data respectively.
[0085] Table 1. Original data table of sucrose ferric acid degradation kinetics
[0086] pH 4.65 4.55 4.46 4.34 4.23 4.16 N 0.583 1.364 5.41 8.68 35.28 61.66 lgN -0.2343 0.1348 0.7332 0.9385 1.5475 1.7900
[0087] Take the logarithm of the measured turbidity value with the base of ten, and do th...
example 1
[0096] Taking the derivative of both sides of formula (d) with respect to pH, we get:
[0097] dN / dpH=-4.1144×10e (-4.1144×pH+18.915) (e)
[0098] Let dN / dpH=-11.4
[0099] Get -4.1144×10e (-4.1144×pH+18.915) = -11.4(f)
[0100] Solve formula (f) to get: pH=4.5775
[0101] That is, the sucrose iron cloud point PP=4.5775
example 2
[0103] dN / dpH=-4.1144×10e (-4.1144×pH+18.915) (e)
[0104] Let dN / dpH=-20
[0105] Get -4.1144×10e (-4.1144×pH+18.915)=-20(g)
[0106] Solve formula (g) to get: pH=4.5182
[0107] That is, the sucrose iron cloud point PP=4.5182
[0108] Secant method example 1
[0109] N=10e(-4.1144×pH+18.915)(d)
[0110] Let N=1, that is, 10e(-4.1144×pH+18.915)=1(h),
[0111] Solving for (h),
[0112] Get pH=4.5970
[0113] That is, the cloud point of iron sucrose PP=4.5970
[0114] Secant method example 2
[0115] N=10e(-4.1144×pH+18.915)(d)
[0116] Let N = 2, i.e. 10e(-4.1144 x pH + 18.915) = 1(i),
[0117] Solving for (i),
[0118] Get pH=4.5243
[0119] That is, the sucrose iron cloud point PP=4.5243
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