The solution of NaOH, water and HCl in the 100-mL was poured into the proper waste container. The solution of 40-mL of water and 5-mL of HCl was prepared again in the same 100-mL beaker and placed under the…
25.0 mL of 0.212 M NaOH is neutralized by 13.6 mL of an HCl solution. The molarity of the NaOH solution is…
10 ml of acid was used for all the three trials to keep a constant. During the first trial it took 3.700 ml of NaOH to titrate the HCl, in the second trial it took 3.750 ml, and in the third trial it took 3.750 ml. These differences in amounts of NaOH required to titrate the acid led to different molarities being calculated for the NaOH. The first being 0.270M and the other to being 0.267M. In order to calculate the total volume of base used, the readings from the burette from before and after titration were subtracted from each other. The average of the three molarities was found to be 0.268M for NaOH (Figure 2).…
The average titre (19.3mL) was used for the volume of sodium hydroxide, whilst the concentration was 0.1 molL-1. 0.00193 moles of sodium hydroxide were used in this experiment.…
A sodium hydroxide solution was prepared. Then 4.2 mL of concentrated 19.1 M NaOH was obtained and mixed with 800 mL dH2O in a 1L bottle.…
1. Moles of NaOH = 0.500 mol/L X 0.010525 L = 5.26 x 10-3 mol…
1. To titrate a hydrochloric acid solution of “unknown” concentration with standardized 0.5M sodium hydroxide.…
The purpose of this experiment is to prepare a standard solution of potassium hydrogen phthalate and use titration to perform an acid/base reaction between the potassium hydrogen phthalate and sodium hydroxide to standardize approximately 0.10 M sodium hydroxide solution. To prepare the Potassium Hydrogen phthalate, a 2.00 grams of KHP was measured to an accurate measurement of 1.980 grams. A total of 100 mL of water was mixed with the KHP solution in the volumetric flask to finally prepare an acidic KHP solution. The molar mass of KHP was calculated and came to be 208.252 grams per mole. To find the moles of KHP, the mass of KHP (1.980 g) was divided by the molar mass of KHP ( 208.252 g) and .00951 moles are in the 1.980 grams of KHP. To find the molarity, the number of moles of KHP (.00951 moles) was divided by the volume of water in liters (.100 L) giving an answer of 0.0951 M.…
In equation 1, the acid is HCl (called hydrochloric acid) and the base is NaOH (called sodium hydroxide). When the acid and base react, they form NaCl (sodium chloride), which is also known as table salt. The titration proceeds until the equivalence point is reached, where the number of moles of acid is equal to the number of moles of base. This point is usually marked by observing a color change in an added indicator.…
25.0 mL of 0.12 mol L−1 standard barium hydroxide solution was titrated with nitric acid. The results are recorded in the table.…
50mL of 0.1 M HCL was obtained in a beaker. A second buret was rinsed with water and then a small amount of HCL, and then filled with HCL. It was then labeled ACID. The initial volume was recorded and then approximately…
In the experiment, we tested a sodium chloride solution. Along with the tested solution, control groups (water and sodium phosphate) were used to be help understand whether or not NaCl was a buffer. Water was the negative control group and sodium phosphate was the positive control group. If NaCl was a buffer than the pH would be stabled as the sodium phosphate buffer. If NaCl was not a buffer than the pH would fluctuate like the negative control, water. During the first trial and prior to the drops of 0.5 M of HCl acid, the pH of sodium chloride was 7.50. After the addition of 5 drops of 0.5 M of HCl, the pH decreased by 4.83 and ended at 2.67 on the pH scale. When comparing the results of the sodium chloride to the control groups, the total pH change of sodium phosphate was only…
The goal of this lab was to correctly prepare a 0.2M solution of NaOH, identify highly acidic household cleaning chemicals, and determine their concentration (molarity) through titrations using the previously prepared 0.2M NaOH solution.…
A sodium hypochlorite (NaOCl) solution was also prepared by mixing 8.00 mL of a stock solution with DI water to obtain 50 mL. The concentration of the NaOCl solution was also determined using the following equation:…
To a 150-mL, 0.1 gram of sodium hydrogen (solid) was measured and added. Drop wise, 1 mL of 1.0 M HCl was slowly added. The characteristics of the reaction were recorded.…