Assuming titration was carried out between sodium hydrogen tetraoxosulphate (vi)(NaHSo) of concentration 12.0gdm-² as solution A and 25cm³ of sodium hydroxide(NaOH) as solution B
What is the volume of the pipette used ?
Well, assuming the titration was done by a professional clown chemist, the volume of the pipette used was probably a clown-sized pipette, capable of holding a whopping 500 liters of liquid! But don't worry, they only used a tiny fraction of it for this experiment.
To find the volume of the pipette used, we need to use the information given in the question. However, it seems like there might be a mistake in the chemical formula provided for sodium hydrogen sulfate. The correct formula for sodium hydrogen sulfate is NaHSO4.
Assuming that the solution of sodium hydrogen sulfate (NaHSO4) has a concentration of 12.0 g/dm³ (grams per cubic decimeter) and the volume of sodium hydroxide solution (NaOH) used is 25 cm³, we can calculate the volume of the pipette used as follows:
1. Convert the concentration of NaHSO4 from g/dm³ to g/cm³:
Concentration in g/dm³ = 12.0 g/dm³
Concentration in g/cm³ = (12.0 g/dm³) / (1000 cm³/dm³) = 0.012 g/cm³
2. Use the equation: concentration (in g/cm³) = mass (in g) / volume (in cm³)
Rearrange the equation to solve for volume:
Volume (in cm³) = mass (in g) / concentration (in g/cm³)
Since the concentration of NaHSO4 is 0.012 g/cm³ and the mass is not given, further information is needed to calculate the volume of the pipette accurately.
To find the volume of the pipette used, we can use the equation:
(concentration of solution A * volume of solution A) = (concentration of solution B * volume of solution B)
Given:
Concentration of solution A (CH3COOH) = 12.0 g/dm³
Volume of solution A = unknown
Concentration of solution B (NaOH) = unknown
Volume of solution B (NaOH) = 25 cm³
We need to find the volume of solution A, which is the unknown variable.
First, we need to establish a balanced chemical equation for the reaction between NaHSo (solution A) and NaOH (solution B) to determine the stoichiometry of the reaction.
The balanced chemical equation for the reaction between NaHSo and NaOH is:
NaHSo + NaOH -> Na2So₄ + H2O
From the balanced equation, we can see that the stoichiometric ratio between NaHSo and NaOH is 1:1. This means that for every 1 mole of NaHSo, we need 1 mole of NaOH.
Next, we can calculate the concentration of NaOH (solution B) using the information given.
The molecular weight of NaOH is 23 + 16 + 1 = 40 g/mol.
Using the concentration formula:
Concentration (g/dm³) = (mass of solute (g) / volume of solution (dm³))
Rearranging the formula:
Mass of solute (g) = Concentration (g/dm³) * Volume of solution (dm³)
Given:
Concentration of solution A = 12.0 g/dm³
Volume of solution A = unknown
Concentration of solution B = unknown
Volume of solution B = 25 cm³
Let's assume the concentration of solution B is labeled as "X" g/dm³.
Using the formula for concentration, we can write:
12.0 g/dm³ * Volume of solution A = X g/dm³ * 25 cm³
To cancel out the units, we need to convert cm³ to dm³. Since 1 dm³ = 1000 cm³, we can divide 25 cm³ by 1000 to get the volume in dm³.
12.0 g/dm³ * Volume of solution A = X g/dm³ * 0.025 dm³
Now, we can solve for Volume of solution A by rearranging the equation:
Volume of solution A = (X g/dm³ * 0.025 dm³) / 12.0 g/dm³
Simplifying the equation, we find:
Volume of solution A = 0.025X / 12.0 dm³
Therefore, the volume of the pipette used is 0.025X / 12.0 dm³.