Document ID : TE1595
Published Date : 10/12/2020
Source : Hach-US

 

Question
What is the purpose of dilution and instructions for performing a dilution?  稀釋的目的為何? 執行稀釋之說明

Summary
Dilution explanation

Answer
Dilutions can be important when dealing with an unknown substance.
A dilution can be performed not only to lower the concentration of the analyte that is being tested, so that it is in range, but also to help eliminate interferences from other substances that may be present in the sample that can artificially alter the analysis.

在處理未知物質時，稀釋可以是重要的一個步驟。
稀釋不僅可以降低被測分析物的濃度使其在範圍值內，還可幫助消除水樣中可能存在的其他物質的干擾，而這些物質可以人為地改變分析。

An analyte is the compound in the sample that is desired to be tested.  An example is if performing a chlorine test, chlorine would be the analyte.

分析物是水樣中需要測試的化合物。 舉個例子，如果進行氯測試，氯將是分析物。

 

An interference is a compound that can add to, or subtract from the result of the analysis. For example, oxidized manganese (Mn +6 ) acts as an interference to the chlorine test. If the Mn +6 is present in the sample it will add to the value of the chlorine test due to the way the it reacts to the chlorine test reagents. By performing a dilution on a sample it may reduce the interfering substance to a point where it no longer interferes with the test.
干擾是可以增加或減少分析結果的化合物。例如，氧化錳 (Mn +6) 會干擾氯測試。 如果 Mn +6 存在於水樣中，由於它與氯測試試劑的反應方式，它將增加氯測試的值。 透過對水樣進行稀釋，可以將干擾物質減少到不再干擾測試的程度。

When performing a dilution the following equation is often cited  在進行稀釋時，經常引用以下等式 :  

C 1 V 1 = C 2 V 2
C 1 is the initial concentration
V 1 is the initial sample volume before dilution
C 2 is the final concentration after dilution
V 2 is the final volume after dilution

Example: To prepare a 2 mg/L Cl 2 standard by dilution of a 50 mg/L Cl 2 primary standard using a 100 mL volumetric flask, below is the calculation to determine the volume of primary standard to add to the volumetric flask.
範例：要透過使用 100 mL 容量瓶(定量瓶)稀釋 50 mg/L Cl 2 一級標準品來製備 2 mg/L Cl 2 標準品，以下是確定添加到容量瓶(定量瓶)中的一級標準品容量的計算方法。

C 1 = 50 mg/L Cl 2
V 1 = Calculated variable in mL
C 2 = 2 mg/L Cl 2
V 2 = 100 mL

50*V 1 =2*100
V 1 =(2*100)/50
V 1 =200/50
V 1 =4 mL

So in this example if 4 mL of the Primary 50 mg/L Cl 2 standard would be added to a 100 mL volumetric flask and then diluent added to the 100 mL graduation, then the solution in the volumetric flask would be 2 mg/L Cl 2 .
因此，在本次範例中，如果將 4 mL 一級 50 mg/L Cl 2 標準品添加到 100 mL 容量瓶(定量瓶)中，然後將稀釋劑添加到 100 mL 刻度，則容量瓶(定量瓶)中的溶液將為 2 mg/L Cl 2 .

This equation can be used to solve for any one of the four variables. For best accuracy use volumetric glassware for volume measurements. 
此等式可用於求解四個變量中的任何一個。 為獲得最佳準確度，請使用體積玻璃器皿進行容量測量。

For more information and additional examples, please see Chemical Analysis the section of the Water Analysis Handbook
更多資訊與範例, 請詳見Hach Water Analysis Handbook (水質分析手冊)中的 Chemical Analysis (化學分析單元).