Molarity Of The Naoh Solution Lab

A higher volume of NaOH will result in more moles of NaOH being added to the HCl, which results in more HCl reacting. This makes the calculated molarity of the HCl be smaller than the actual molarity of the HCl.

The results showed the molarity of the NaOH solution. This experiment was completed twice and a new average molarity

In 2 and 7 I added 50 mL of .1 M NaCl. I added sodium acetate to the rest of the beakers: 1 gram to 3 and 8, 5 grams to 4 and 9, and 10 grams to 5 and 10. I then filled the beakers that contained the solid sodium acetate with 50 ml of .10 M acetic acid. Specifics can be found on page 84 of the lab manual. Though the lab manual instructed to use a pipet, we did not have an accurate 1 mL pipet or a graduated pipet, so we instead prepared two graduated burets with 1 M Sodium Hydroxide and 1 M hydrochloric acid. Using a standardized pH probe with a Lab Pro to measure changes in pH, we added 1 mL of HCl at a time and recorded the changes. The same was done for the NaOH.

In this lab, the molar mass of a volatile liquid is determined based on its physical properties in the vapor state. In order to calculate the molar mass, the mass, temperature, pressure, and volume is measured independently and then converted to the correct units. Sample C was obtained at the beginning of the experiment, which was later informed to be ethanol. Based on the calculations made, the molar mass of the volatile liquid was 95.9 g/mol. However, compared to the known value of 46.1 g of ethanol, the value measured had a 108% error. Unfortunately, this was a very big percent error and may have been caused by incorrectly measuring the volume of the gas. Using the ideal gas law, the molar mass of a volatile compound was calculated in order

Chemistry 102 is the study of kinetics – equilibrium constant. When it comes to the study of acid-base, equilibrium constant plays an important role that tells how much of the H+ ion will be released into the solution. In this lab, the method of titrimetry was performed to determine the equivalent mass and dissociation constant of an unknown weak monoprotic acid. For a monoprotic acid, it is known that pH = pKa + log (Base/Acid). When a solution has the same amount of conjugate base and bronsted lowry acid, log (Base/Acid) = 0 and pH = pKa. By recording the pH value throughout the titration process and determining the pH at half- equivalence point, the value of Ka can be easily calculated. In this experiment, the standardized NaOH solution has a concentration of 0.09834 M. The satisfactory sample size of known B was 0.2117 g. The average equivalent mass of the unknown sample was found to be 85.01 g, pKa was found to be 4.69, which was also its pH at half-equivalence point and Ka was found to be 2.0439×〖10〗^(-5). The error was 1.255% for equivalent mass and 0.11% for Ka. In other word, the experiment was very precise and accurate; the identity of the unknown sample was determined to be trans-crotonic by the method of titrimetry.

Determining the Molar Volume of a Gas Anita Lau Partner: Anthony Yuen Ms B. IDC4U 24 April, 2015 Purpose: In this experiment, the molar volume ( the volume occupied by one mole of a gas) of hydrogen gas at standard temperature and pressure is measured. According to Avogadro's Law, at the same temperature and pressure, equal volumes of gases contain the same number of molecules. Therefore the volume of any given gas must be proportional to the number of moles of molecules present when the temperature and pressure are constant.

In 1909 S.P.L. Sorensen published a paper in Biochem Z in which he discussed the effect of H1+ ions on the activity of enzymes. In the paper he invented the term pH to describe this effect and defined it as the -log[H1+ ]. In 1924 Sorensen realized that the pH of a solution is a function of the "activity" of the H1+ ion not the concentration and published a second paper on the subject. A better definition would be pH=-log[aH1+ ], where aH1+ denotes the activity of the H1+ ion. The activity of an ion is a function of many variables of which concentration is one. It is unfortunate that chemistry texts use a definition for pH that has been obsolete for over 50 years.

The samples were synthesized from a synthesis solution by dissolving 7.98 g sodium hydroxide pellets (A.R) and 11.01, 8.01, 6.79, 2.73, 1.64, 0.9 g of aluminum sulphate, aluminum chloride, aluminum isopropoxide, sodium aluminate, alumina and aluminum metal (Aldrich), respectively in 69.5 g deionized water in a beaker. The mixtures in the beaker were thoroughly mixed and a 50 g Ludox AS30 colloidal silica (Aldrich) was slowly added to the above solution under stirring at high speed. The molar composition of the resulting synthesis gel was 12Na2O: 100SiO2:2Al2O3: 500H2O. Prior to being transferred to a Teflon-lined stainless steel autoclave, the above synthesis solution was aged for 20 hr at room temperature and then hydrothermally treated

Experiment to investigate the amount of sodium hydroxide needed to neutralize the solution of vinegar

By using a catalyst to break down the Hydrogen Dioxide into oxygen and water, my partner and I were able to determine the molar volume of oxygen from the volume we produced in the experiment. The chemical equation for the reaction was:

Foor this you simply have to take a look at the processed data table and graph. There is a clear specific trend, visibly recognizable as well as mathematically (a trend line indicates it), that is, as the molar concentration (solute) of salt increased in the solution where potatoes were placed, the mass of potatoes decreased. Now, from the molar concentrations of salt solute that we tested (0.0M, 0.2M, 0.4M, 0.6M, 0.8M, 1.0M) we cannot decipher with certainty and accuracy what is the molar concentration of potato cells. However, we can estimate, and judging by the a decrease in percent mass from 21.10% grams (0.0M) to 12.13% (0.2M) and followed by a constant decrease in percent change in mass, it can be said that the salt concentration of a potato cell is between 0.0M and 0.2M. This is because these molar concentrations are the one that show the decrease in mass of the potato. When there’s a decrease in mass, this practically means that the water inside the cells is leaving because of diffusion to areas of lower concentration.

In Part 1, ~0.30g of KHP is dissolved in 25.00 mL of water, and titrated with NaOH. The volume of NaOH added to reach the endpoint is recorded and corrected (Table 1). The number of moles of NaOH reacted in trial 1 and trial 2 are 0.001534 mol and 0.001433 mol (Equation 1). The concentration of NaOH in trial 1 and trial 2 are 0.07331 M and 0.06628 M (Equation2). The average concentration of NaOH is 0.06979 M (Equation 3). The average deviation is 0.003515 M (Equation 4). In Part 2, ~0.13g of Na2CO3 is dissolved in 25.00 mL of water, and titrated with HCl. The volume of HCl added to reach the endpoint is recorded and corrected (Table 2). The number of moles of HCl reacted in trial 1 and trial 2 are 0.001225mol and 0.001253mol (Equation 5). The concentration of HCl in trial 1 and trial 2 are 0.03298M and 0. 0.03477M (Equation 6). The average concentration of HCl is

The purpose of this experiment is to determine the molar mass of an unknown volatile organic liquid, using methods such as, the Dumas method, and then state its identity.

Trial two started with 0 mL of NaOH and ended at 17.3 mL. On trial three NaOH started at 0 mL and it ended at 21.5 mL. The molarity of HCL was 0.1 M and its volume was 20 mL. The average of sodium hydroxide used was 19 mL. Using the dilution equation, 0.1 M of hydrochloric acid multiplied by 20 mL of hydrochloric acid is 2. Taking 2 divided by 19 mL of sodium hydroxide gives 0.105 M as the molarity for NaOH. The calculated unknown molarity of NaOH was 0.105 M. This compares to the known theoretical molarity of NaOH which was 0.1 M. The molarity of sodium hydroxide from the experiment is close to its theoretical hydroxide meaning the results of the experiment were accurate.

The purpose of this experiment was to determine how much KMnO4 was needed to titrate approximately 1 mL of an Unknow X101 concentrated solution of Oxalic Acid. A standardized KMnO4 solution was used on a known solution of Oxalic acid to help determine the unknown percent oxalic acid in unknown X101. The unknown sample for this experiment was sample x101 which theoretically was a % Oxalic Acid dehydrate sample but, the average of all three trials determined it to be a 6.7% percent Oxalic acid.