been standardized. You might think that you could just weigh 4.000g of solid NaOH and dissolve it in enough water to make 1.000 L of solution. Equipment/Materials: Calculate the molar concentration of the sodium hydroxide solution using your data for each run. Standardizing NaOH(aq) Potassium hydrogen phthalate (KHP, KC 8 H 5 O 4) is a solid, monoprotic acid. 37 - 24,4 = 12,4 into the acid HCL The reaction between solutions of HCl and NaOH is illustrated by Equation 1. Suppose that you needed a 0.1000 M solution of NaOH to do a titration. Calculation N 1 V 1 = N 2 V 2 N 2 =Normality of oxalic acid N 1 =Normality of NaOH V 2 =Volume of Oxalic acid V 1 =Volume of NaOH N 1 =N 2 V 2 /V 1. Before you can use the NaOH(aq) to standardize your HCl(aq), you will have to standardize the NaOH(aq) using the primary solid acid standard, potassium hydrogen phthalate. By: Juno Kim Introduction In this experiment the concentration of potassium hydrogen phthalate (KHP) in an unknown sample was determined through volumetric analysis. 4.000 g is after all 0.1000 mole of NaOH, so dissolving it in one litre of water should produce a one litre solution. Weight out 0.4g of KHP 2. Try it risk-free for 30 days Try it risk-free Ask a question. STOPPER AND SAVE YOUR NaOH SOLUTION FOR USE IN EXPERIMENT 12B !!!!! NaOH of unknow concentration. This is also the case for the standardization of NaOH with standard HCl. Top it up to the mark. In this experiment, standardization of a NaOH solution will be carried out either using KHP as the primary standard or by using a standard HCl solution of known concentration. This sample calculation uses the … PROCEDURE (B): TITRATION OF STANDARDIZED NaOH AGAINST 12M HCL (1) Prepare 500ml of about 0.1M HCL from the concentrated HCL available in the laboratory by pipetting 4.2ml of the acid solution into a graduated cylinder. We ended with 24,4 ml of NaOH. Calculate the average molar concentration of the sodium hydroxide solution. CONCLUSION. Become a member and unlock all Study Answers. Calculations : 24,4 ml - 12 ml = 12,4 ml ⇒ Neutral. THIS IS THE VALUE THAT YOU WILL USE IN EXPERIMENT 12B. Therefore, the equivalent weight of crystalline oxalic acid = 63. Oxalic acid has the formula C 2 H 2 O 4 (HOOC-COOH) so it is diprotic . E. CALCULATION 1. Average NaOH molarity calculated to be 0.09515 M. Titrations performed by Yu-Ting Tseng and taken for comparison The standard deviation for both data sets was calculated by the method shown below using Equation 1. (2) Again fill the burette with the standardized NaOH solution to the zero mark. From the above experiment it was evident that sodium hydroxide can be effectively standardized by using oxalic acid. 3. 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