## Sunday, May 31, 2015

### Volumetric Analysis Problems, 5

Category: Chemical Engineering Math

"Published in Newark, California, USA"

A 5.8734 gram sample of beef was analyzed for N content and the liberated ammonia was collected in a 50.00 mL of 0.4691 M HCl and a 12.55 mL back titration with 0.0256 M NaOH was required. Calculate the percentage protein in the beef sample.

Solution:

The given word problem is about volumetric analysis in which the titration of a sample and a solution is involved. By Kjeldahl method, we can analyze the amount of protein present in any organic samples.

If liberated ammonia is titrated with HCl solution, then the chemical reaction is

and the solution is back titrated with NaOH solution is

The amount of liberated ammonia in a sample is

By gravimetric analysis, the amount of nitrogen present in a sample is

Therefore, the percent of nitrogen or protein in a sample of beef is

## Saturday, May 30, 2015

### Volumetric Analysis Problems, 4

Category: Chemical Engineering Math

"Published in Newark, California, USA"

A 758 mg sample of full cream milk was analyzed by the Kjeldahl method; 38.61 mL of 0.1078 M HCl were required to titrate the liberated ammonia. Calculate the % N in the sample.

Solution:

The given word problem is about volumetric analysis in which the titration of a sample and a solution is involved. By Kjeldahl method, we can analyze the amount of protein present in any organic samples.

If liberated ammonia is titrated with HCl solution, then the chemical reaction is

The amount of liberated ammonia in a sample is

By gravimetric analysis, the amount of nitrogen present in a sample is

Therefore, the percent of nitrogen in a sample of full cream milk is

## Friday, May 29, 2015

### Volumetric Analysis Problems, 3

Category: Chemical Engineering Math

"Published in Newark, California, USA"

A sample consisting of Na2CO3, NaHCO3 and inert matter weighs 1.179 grams. It is titrated with 0.1 N HCl with phenolphthalein as the indicator, and the solution became colorless after the addition of 24.00 mL. Another duplicate sample was titrated with HCl using methyl orange as indicator. It required 50.25 mL of the acid for the color change from yellow to red. What are the percentage of Na2CO3 and NaHCO3 in the sample?

Solution:

Since a given sample consists of Na2CO3 and NaHCO3, then we have to use two indicators like phenolphthalein and methyl orange. The change of color of an indicator depends with the pH of the end point. The pH of the end point of phenolphthalein is usually 7. It is changed from pink to colorless. The pH of the end point of methyl orange is less than 4 or 5. It is changed from yellow to red color.

In the given problem, the titration method is done twice in a separate method using equal duplicate samples. This method is applicable if the mixture consists of Na2CO3 and NaHCO3

In the first titration using phenolphthalein as the indicator, then we can solve for the amount of NaHCO3 in a sample. The end point must be a colorless solution. The chemical reaction for the first titration is

Hence, the weight of NaHCO3 in the first titration is

In the second titration using methyl orange as the indicator, then we can solve for the amount of Na2CO3 in a sample. The end point must be a red color solution. The chemical reaction for the second titration is

Hence, the weight of NaHCO3 in the second titration is

Next, let's check if NaHCO3 is present in the sample as follows

Since the right side of the equation is greater than the left side of the equation, then the given sample has NaHCO3. If both sides of the equation are equal, then only Na2CO3 is present in the sample

Hence, the weight of NaHCO3 in the sample is

Therefore, the percent of NaHCO3 in the sample is

The weight of Na2CO3 in the sample using the chemical equations above is

Therefore, the percent of Na2CO3 in the sample is

## Thursday, May 28, 2015

### Volumetric Analysis Problems, 2

Category: Chemical Engineering Math

"Published in Newark, California, USA"

A sample consisting of Na2CO3, NaOH, and inert matter weighs 1.179 grams. It is titrated with 0.2239 M HCl with phenolphthalein as the indicator, and the solution became colorless after the addition of 45.62 mL. Methyl orange is then added and 12.85 mL more of the acid are needed for the color change. What are the percent of Na2CO3 and NaOH in the sample?

Solution:

Since a given sample consists of Na2CO3 and NaOH, then we have to use two indicators like phenolphthalein and methyl orange. The change of color of an indicator depends with the pH of the end point. The pH of the end point of phenolphthalein is usually 7. It is changed from pink to colorless. The pH of the end point of methyl orange is less than 4 or 5. It is changed from yellow to red color.

If a sample is added with phenolphthalein and titrated with HCl solution, then the solution changed its color from pink to colorless. The chemical reaction for the first titration is

Sodium hydroxide is neutralized completely and the endpoint is colorless. The weight of NaOH in a sample is

Since Na2CO3 is converted into NaHCO3 and NaHCO3 is not yet a stable salt, then we need to do the titration again by adding methyl orange and then titrate with HCl solution until the color is changed from yellow to red. The chemical reaction for the second titration is

The weight of NaHCO3 is

By gravimetric analysis and from the chemical equations above, the weight of Na2CO3 is

Therefore, the percent of Na2CO3 in a sample is

and the percent of NaOH in a sample is