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11 Nov 2019

I need help setting up the last two parts of this problem. I do not want the answer, just to understand how to set it up and why.

Sodium hydrogen carbonate, NaHCO3 is a base used in several commercial antacids. The overall reaction with HCl is: NaHCO3(aq) + HCl(aq) â NaCl (aq) + H20 (l) + CO2 (g) 1. A tablet, which had a mass of 1.213 g, was pulverized and 0.613 g were dissolved in 25.00 mL of 0.220 M HCl. The titration of this solution with 0.1615 M NaOH required 26.30 mL to reach the phenolphthalein end.point. Determine the moles of HCl that were added to the sample analyze I found this answer to be.00550 mol 2. For the previous problem, determine the moles of HCl that reacted with the NaOH. I found this answer to be 0.00425 mol These last two are where I am stuck and don't know how to set up the problem. I don't want the answer, I just want help understanding how to set it up. 3. Determine the grams of NaHCO3 in the sample analyzed. 4. Determine the mass of NaHCO3 in the tablet.

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Reid WolffLv2

10 Mar 2019

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So I did a titration lab where were reacted various amounts of Mg with 10mL of HCL and then neutralized the raction with NaOH. I have my table of values for Mass of Mg (g), Initial Volume NaOH (mL), Final Volume NaOH (mL), End Point Volume (mL), Limiting Reagent (Mg or HCL). There is a Titration table I need to fill in that asks for Moles of Mg (mol), Moles of NaOH (mol) and Moles of H⁺ reacting with Mg (mol). I have the Moles of Mg calculated and I also have Moles of NaOH (mol) calculated but how do I go about finding Moles of H⁺ reacting with Mg. The manual says I can use: Moles of H⁺ total - moles of H⁺ excess = moles H⁺ reacting with metal. Where could I obtain those numbers.

I had in my notes to take Moles of Mg - Moles of NaOH and that gives the value but the problem with I am getting with that method is that if my limiting reagent is Mg then my Moles of H⁺ calculation ends up being a negative number.

My formula for calculating Moles of Mg is (Mass of Mg starting) * (1Mol/24.31g) = Moles of Mg

My formula for Calculating Moles of NaOh is: (End Point Volume)*(1L/1000mL)*(.3M HCL/1L) = Moles of NaOH

Thanks!

Some examples of my numbers would be

Moles of Mg = .001, Moles of NaOH = 8.58*10^-3 so that gives me a -7.6*10^-3 but I don't know if that is correct. Any help?

I have a question that is based on this experiment, but I really don't understand how to extrapolate the data to answer the questions that asked, and I don't think I can answer the question. Can someone help me?

The Experiement

Find the actual mass of ASA (C₉H₈O₄) (K_a = 3.2 x 10^-4) in one tablet using a titration with standardized 0.100 mol/L NaOH_{(Aq). Â}NOTE: it states that an actual ASA tablet has an approximate mass of 500mg (0.5g), and the tablet is not pure ASA.

Step 1:Use 0.500g of a crushed ASA tablet

Step 2: Add 10mL of distilled water to a flask to dissolve the ASA

Step 3: Add an indicator to the flask

Step 4: Fill a burette with standardized NaOH_(aq) and place the flash under the stopcock of the burette

Step 6: Add the base to the flask in small amounts. Swirl the flask between additions of base.

Step 7: Stop adding base when the indicator has changed colour.

Step 8: Record the volume of base used

Step 9: Repeat the procedure as many times as needed to have three consistent volumes.

The data for observations of four titratons was:

Titration trial #	Volume of NaOH_(aq) used in mL
1	18.1
2	17.9
3	12.4
4	18.0

The question asks me to Calculate the number of moles and the mass of ASA in one tablet. Determine the percentage of the original mass of tablet that was actually ASA. This would be the percent purity of the tablet and would give an indication of the amount of filler that was present in the tablet.

Federal regulations set an upper limit of 50 parts per million (ppm) of NH₃ in the air in a work environment [that is, 50 molecules of NH₃(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00 × 10² mL of 0.0105 M HCl. The NH₃ reacts with HCl according to:

NH₃(aq) + HCl(aq) NH₄Cl(aq)

After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. (a) How many grams of NH₃ were drawn into the acid solution? (b) How many ppm of NH₃ were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 g/ mol under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?

After bubbling the air through the acid solution, remaining amount of acid is calculated from titration with NaOH:

HCl(aq) + NaOH(aq) NaCl(aq) + H₂O(l)

Mol of HCl in excess:

The amount of HCl consumed for neutralization of ammonia from air is:

The same amount of NH₃ was also present in a sample, as ammonia and HCl react in 1:1 ratio.

Molar mass of NH₃ is 17.0 g/mol, and the mass drawn into acid is:

b) The volume of air passed through the HCL solution is:

Number of moles of air:

The concentration of ammonia in the air, in ppm, is:

c) The manufacturer is not in compliance with regulations since it releases more than 50 ppm of ammonia in the air.