A bottle of commercial hydrochloric acid solution So is labeled 37% composition by mass, density 1.19 kg/L.
Calculate the concentration of solution So.
4.2 ml of So are diluted to 500ml solution called S. We titrate S with 200 ml of KOH solution of 3×10^-3 mol/l placed in a beaker.
Determine the equivalence point and then find the concentration of solution S.
Deduce the concentration of solution So and its% composition by mass. Compare It to the label
i have no idea
To calculate the concentration of solution So, we need to use the given information:
1. The bottle of solution So is labeled as having a 37% composition by mass.
2. The density of solution So is 1.19 kg/L.
First, let's convert the density of the solution from kg/L to g/mL:
1 kg = 1000 g
1 L = 1000 mL
Therefore, the density of the solution So is 1.19 kg/L = 1.19 g/mL.
Now, we can calculate the concentration of solution So. The concentration of a solution is usually given in terms of moles per liter (mol/L) or molarity.
To find the molarity of solution So, we need to calculate the number of moles of the solute (hydrochloric acid) in one liter of the solution.
To do this, we need to consider the molecular weight (molar mass) of HCl, which is 36.46 g/mol.
The formula to calculate the molarity of a solution is:
Molarity (M) = (mass of solute (g)) / (molar mass (g/mol)) / (volume of solution (L))
The mass of solute (hydrochloric acid) in one liter (1000 mL) of solution So can be calculated as follows:
mass of solute (g) = (37% composition) * (density of solution) * (volume of solution)
mass of solute (g) = (0.37) * (1.19 g/mL) * (1000 mL)
Now we can calculate the molarity (concentration) of solution So:
Molarity (M) = (mass of hydrochloric acid (g)) / (molar mass of HCl (g/mol)) / (volume of solution (L))
Molarity (M) = ((0.37) * (1.19 g/mL) * (1000 mL)) / (36.46 g/mol) / (1 L)
After calculating this, we will obtain the concentration (molarity) of solution So.
For the second part of the question:
4.2 mL of solution So is diluted to 500 mL, resulting in a solution called S.
This means that solution S has been diluted by a factor of (500 mL / 4.2 mL) = 119.05 times.
Next, we titrate solution S with 200 mL of a known concentration KOH solution (3×10^-3 mol/L). The titration is performed until we reach the equivalence point.
The equivalence point is the point in a titration where the stoichiometrically equivalent amounts of acid and base have been mixed together. In this case, it is the point where all the hydrochloric acid in solution S reacts with the potassium hydroxide (KOH).
To find the concentration of solution S, we can use the following equation which relates the moles of acid and base at the equivalence point:
(moles of acid) * (volume of acid solution) = (moles of base) * (volume of base solution)
In this equation, the moles of base is equal to the concentration of KOH solution (3×10^-3 mol/L) times the volume of KOH solution (200 mL).
We already know that the volume of the acid solution is 4.2 mL.
By plugging in these values and rearranging the equation, we can solve for the moles of acid, and subsequently, the concentration of solution S.
Once we have the concentration of solution S, we can calculate the concentration of solution So.
To do this, we need to use the dilution factor (119.05) and relate the concentrations of solution So and solution S:
Concentration of solution S = (Concentration of solution So) * (dilution factor)
Finally, we need to compare the resulting concentration of solution So with the labeled concentration of 37% by mass.