ch group will make up 1-2 solutions and pool the results. Your instructor will g

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ch group will make up 1-2 solutions and pool the results. Your instructor will get an average of each s measurements, Students will compare the groups' averages versus that calculated for the theoretical 1. Using a 100 mL volumetric flask, make up one or more 0.10 M solution of the salts in the Lab ph. Equipment and Chemicals Section as assigned by your instructor. Label the flask 2. Calibrate your pH meter using pH 4 and 7 buffers 3. Place approximately 40 mL in a 50 mL beaker, label the beaker and measure the pH. Share this solution with other groups. 4.0, Hel 4. Measure the pH of a sample of distilled water. .7 5. Record the results in your notebook. At the completion of the above measurements: 6. Use your distilled water in step 3, add one drop of 6M HCl and remeasure the pH. 3.(2 7. Taking the original solution put approximately 40 mL in a 50 mL beaker and measure the pH. Discussion Questions (to be discussed in the conclusion of your lab report) 1. Describe how you made your 0.10 M salt. Be specific. 2. Besides confirming that the water does not hydrolyze to any great extent, why was the pH of distilled water tested? 3. If you need the Ks for a particular basic salt hydrolysis and you can only find the K, how can you determine the Kb? 4. Describe your observations/results from Procedure steps 5 & 6? 5. What are possible sources of error in this e References: and pH of salt solutions.doc

Explanation / Answer

Since the complete experiment hasn't been described in the image, I am assuming that this experiment has to do with "measurement" of pH of solutions of salts & acids.

1. Assuming that the salt that you’ve taken for preparing its solution be NH4Cl (my guess since in Step 3 you’ve obtained a pH of 4.06 for the given salt solution, which suggests a salt in which the cation is undergoing hydrolysis), then a 100 ml solution of the given salt of concentration 0.1 M can be prepared by dissolving 0.01 moles of NH4Cl which is equal to 0.01 x 52.5 grams = 0.525 grams of NH4Cl.

2. The pH of pure distilled water is 7, which suggests that it is nether acidic nor alkaline. In the process of distillation, water is boiled & then condensed which converts vapours into liquid. But immediately after distillation, distilled water starts absorbing carbon dioxide from atmosphere and slowly forms carbonic acid. This turns distilled water slightly acidic and that’s the reason you have found the pH of distilled water to be 6.76

3. The pH of a solution basic salt (for example NH4Cl) can be calculated using the following equation:

pH = 7 – ½ [pKb + log10C]

where Kb = dissociation constant of NH4OH & C = concentration of salt

However if the intent is to find Ka then after calculating pKb from the above equation by incorporating the value of pH measured at a given value of concentration, we’ll subtract that value from 14 by using the following equation.

pKa + pKb = 14

Now by taking the antilog of the value of pKa obtained using the above equation we can calculate the value of Ka.

4. When a drop of 6M HCl is added to 40 ml of salt solution common ion effect would be observed

Let’s assume 1 drop of HCl = 0.05 ml so the volume of HCl added is 5 x 10-5 litres.

Number of moles of HCl added = molarity x volume = 6 x 5 x 10-5 moles = 3 x10-4 moles

Total volume of solution + volume of salt solution + volume of HCl added = 0.04 litre + 0.00005 litre = 0.04005 litres

Molarity of new HCl solution = 3 x10-4 moles/0.04005 litres = 0.00749 M

pH of this solution = -log10[H+] = 2.12

[We are considering common ion effect to perate in this scenario since H+ given by strong electrolyte HCl would complete suppress the equilibrium that would result from the reaction of NH4+ with water which also gives H+ ions]

5. The possible sources of errors in this experiment are:

a) incorrect measurement of weights and volumes leading to incorrect values of concentration. This will affect the pH calculation as well.

b) Adding more than 1 drop of HCl in step 3.

c) Misreading the marking on burette or volumetric flask. If the measurement is looked at from an angle, a mistake would creep in. The flask or the burette should be kept at eye level. A solution forms a concave curve and the bottom of this curve should be used to measure the volume.

d) Some volume of solution gets accidently spilled out, leads to an error.

Hope this helps. Kindly, rate the answer and also drop in your valuable feedback.

Please post your queries related to the solution (if any).

THANKS

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