The label of a bottle containing a solution So of hydrochloric acid carries the following indications :hydrochloric acid of density 1190g/l percentage by mass 37%. We introduce a volume V=4.2ml of So in a volumetric flask of volume 500ml containing about 100ml of distilled water and we complete the volume with distilled water until the linemark. Te new solution is noted S.

Determine the concentration of the solution S prepared.

In order to verify this concentration we titrate S by using a sloution B of potassium hydroxide of concentration 4×10^-2 mol/l. In 20ml of this solution we pour Vs ml of the solution S and we measure the pH after each addition.
V at equivalence 8.1 ml at pH 7.
Deduce theoretically the pH the solution obtained after the addition of 7mlof S to the 20ml of the basic solution .

To determine the concentration of the solution S prepared, we need to use the given information about the original solution So and the dilution process.

1. Calculate the mass of hydrochloric acid in the original solution So:
The density of the So solution is given as 1190 g/L.
Since the percentage by mass of hydrochloric acid is 37%, we can calculate the mass of hydrochloric acid in 1 liter of the solution as:
Mass[HCl] = Density × Volume × Percentage
Mass[HCl] = 1190 g/L × 1 L × 0.37 = 439.3 g

2. Calculate the mass of hydrochloric acid in the volume V (4.2 mL) of solution So used for dilution:
Mass[HCl] in V of So = Mass[HCl] in 1 L of So × (V/1000)
Mass[HCl] in V of So = 439.3 g × (4.2/1000) = 1.84 g

3. Calculate the total volume of the diluted solution S:
The volumetric flask has a volume of 500 mL, and we added So (4.2 mL) and distilled water (100 mL) initially.
Total volume of S = 500 mL - 4.2 mL - 100 mL = 395.8 mL

4. Calculate the concentration of solution S:
Concentration of S = (Mass of solute) / (Volume of solution)
Concentration of S = (1.84 g) / (395.8 mL) = 0.00464 g/mL = 4.64 g/L

Therefore, the concentration of the solution S prepared is 4.64 g/L.

Now, let's move on to the second part of the question regarding pH calculations.

5. We are titrating solution S with a potassium hydroxide solution of concentration 4 × 10^-2 mol/L.
During titration, the pH is measured after each addition of Vs mL of solution S.

6. Given that at the equivalence point (when the acid and base react completely) for the titration, the volume of solution S used is 8.1 mL, and the pH is 7.

To determine the pH after the addition of 7 mL of solution S to 20 mL of the basic solution, we need to use the concept of the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is given by:
pH = pKa + log (base/acid)

7. In this case, since we are adding the acid solution (solution S) to the basic solution (potassium hydroxide), the concentration of the base will be the initial concentration of potassium hydroxide (4 × 10^-2 mol/L), and the concentration of the acid will be the concentration of hydrochloric acid in the final solution after adding 7 mL of S.

First, calculate the moles of hydrochloric acid in 7 mL of solution S:
Moles[HCl] = (Mass[HCl]) / (Molar mass[HCl])
Molar mass[HCl] = 36.46 g/mol (molar mass of HCl)
Moles[HCl] = (1.84 g) / (36.46 g/mol) = 0.0505 mol

8. Calculate the new concentration of hydrochloric acid in the final solution:
Concentration[HCl] = (Moles[HCl]) / (Volume of solution)
Concentration[HCl] = (0.0505 mol) / (27 mL)
Concentration[HCl] = 0.001873 mol/mL

9. Calculate the ratio of base to acid (potassium hydroxide to hydrochloric acid):
Ratio = (Concentration of base) / (Concentration of acid)
Ratio = (4 × 10^-2 mol/L) / (0.001873 mol/mL)
Ratio = 21.37

10. Use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log (Ratio)
pH = 7 (given) + log (21.37)
pH = 7 + 1.33 = 8.33

Therefore, theoretically, the pH of the solution obtained after adding 7 mL of solution S to 20 mL of the basic solution is approximately 8.33.

To determine the concentration of solution S, we need to calculate the amount of hydrochloric acid present in the solution.

1. Calculate the mass of hydrochloric acid in the original solution So:
Mass = Density * Volume
Mass = 1190 g/L * 4.2 mL
Mass = 4989 g

2. Calculate the mass of hydrochloric acid in the new solution S (after dilution):
Since the original solution is diluted to 500 mL, the mass of the acid remains the same.
Mass = 4989 g

3. Calculate the concentration of solution S:
Concentration = Mass of HCl / Volume of S
Volume of S = 500 mL
Concentration = 4989 g / 500 mL
Concentration = 9.978 g/L

Therefore, the concentration of solution S is approximately 9.978 g/L or 9.978%.

Now, let's move on to the second part of the question.

To deduce the pH of the solution obtained after adding 7 mL of S to 20 mL of basic solution B, we can use the concept of neutralization reaction between acid and base.

1. First, calculate the number of moles of hydrochloric acid added:
Amount of HCl = Concentration * Volume
Amount of HCl = 9.978 g/L * (7 mL / 1000 mL) (convert mL to L)
Amount of HCl = 0.06985 g

2. Since hydrochloric acid and potassium hydroxide react in a 1:1 ratio, the number of moles of KOH used in the reaction is also 0.06985 mol.

3. Calculate the new volume after adding S to B:
Total volume = Initial volume of B + Volume of S
Total volume = 20 mL + 7 mL
Total volume = 27 mL

4. Calculate the resulting concentration of KOH in the new solution:
Concentration of KOH = Amount of KOH / Total volume
Concentration of KOH = 0.06985 mol / 0.027 L (convert mL to L)
Concentration of KOH = 2.58 mol/L

5. Finally, deduce the pH of the solution:
The reaction between KOH and HCl produces water (H2O) and a salt (KCl). Therefore, the resulting solution is neutral, and the pH is 7.

Hence, after adding 7 mL of solution S to 20 mL of basic solution B, the pH of the new solution is theoretically 7.