Solution A in H2So4 containing
0.085mol /dm^3. solution B is
KoH. Titrate using methyl orange
as an indicator. calculate:
A.concentration of A in g/dm^3
B.Concentration of B In mol/dm^3
C.Concentration of B in g/dm^3
take Va=27.30cm^3,Vb=25.0
0cm^3?
A. 0.085 mols/dm^3 x 98 g/mol = ? grams/dm^3
B. volume H2SO44 = 27.30 cc. volume KOH = 25.00 cc
Titration to the M.O. end point means both Hs of the H2SO4 were titrated
H2SO4 + 2KOH ==> K2SO4 + 2H2O
mols H2SO4 = M x L = 0.085 x 0.02730 = about 0.0023 but that a close estimate and you should redo it more accurately
mols KOH = 2 x mols H2SO4 = about 0.0046
M KOH = mols KOH/L KOH = about 0.0046/0.02500 = about 0.18 Mol/dm^3
C. KOH = about 0.18 mol/dm^3
0.18 mols/dm^3 x molar mass KOH = g/dm^3
To solve this problem, we need to use the concept of stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction.
Let's break down each part of the problem:
A. Concentration of A in g/dm^3:
To calculate the concentration of A in grams per cubic decimeter (g/dm^3), we need to know the molar mass of A. Since it is not given in the question, we cannot directly calculate it. We would need additional information about the compound A.
B. Concentration of B in mol/dm^3:
Given the volume (Va = 27.30 cm^3) and concentration (Ca = 0.085 mol/dm^3) of solution A, we can determine the number of moles of A using the formula:
Na = Ca * Va / 1000 dm^3
Na = (0.085 mol/dm^3) * (27.30 cm^3) / (1000 cm^3/dm^3) = 0.00232 moles
C. Concentration of B in g/dm^3:
To calculate the concentration of B in grams per cubic decimeter (g/dm^3), we need to balance the reaction between A and B and determine the stoichiometric ratio.
H2SO4 + 2KOH -> K2SO4 + 2H2O
From the balanced equation, we can see that the stoichiometric ratio of H2SO4 to KOH is 1:2. This means that for every one mole of H2SO4, we need 2 moles of KOH.
Using the volume (Vb = 25.00 cm^3) and concentration (Cb) of solution B, we can determine the number of moles of B reacted with A using the formula:
Nb = (2/1) * Na
Nb = 2 * 0.00232 moles = 0.00464 moles
Now, to calculate the concentration of B in grams per cubic decimeter (g/dm^3), we need to know the molar mass of B. Without additional information about the compound B, we cannot directly calculate its concentration in grams per cubic decimeter.
In summary:
A. We need more information about the compound A to calculate its concentration in grams per cubic decimeter (g/dm^3).
B. The concentration of B in mol/dm^3 is 0.00232 moles.
C. We need more information about the compound B to calculate its concentration in grams per cubic decimeter (g/dm^3).
To calculate the concentration of A in g/dm^3 and the concentration of B in mol/dm^3 and g/dm^3, we will first determine the number of moles of A and B used in the titration.
Given:
Concentration of A (H2So4) = 0.085 mol/dm^3
Volume of A used (Va) = 27.30 cm^3
Volume of B used (Vb) = 25.00 cm^3
A. Concentration of A in g/dm^3:
To find the concentration of A in g/dm^3, we need to convert the concentration from mol/dm^3 to g/dm^3.
Molar mass of H2So4 = (2 x 1) + (1 x 32) + (4 x 16) = 2 + 32 + 64 = 98 g/mol
Number of moles of A = Concentration of A x Volume of A
Number of moles of A = 0.085 mol/dm^3 x (27.30 cm^3/1000 cm^3/dm^3)
Number of moles of A = 0.0023205 mol
Concentration of A in g/dm^3 = Number of moles of A x Molar mass of A
Concentration of A in g/dm^3 = 0.0023205 mol x 98 g/mol
Concentration of A in g/dm^3 = 0.2273 g/dm^3
Therefore, the concentration of A in g/dm^3 is 0.2273 g/dm^3.
B. Concentration of B in mol/dm^3:
To find the concentration of B in mol/dm^3, we need to use the balanced equation of the reaction between A (H2So4) and B (KoH).
The balanced equation is:
H2So4 + 2KoH -> K2So4 + 2H2O
From the balanced equation, we can see that the stoichiometry between H2So4 and KoH is 1:2. This means that 1 mole of H2So4 reacts with 2 moles of KoH.
Number of moles of KoH = (Number of moles of A) / 2
Number of moles of KoH = 0.0023205 mol / 2
Number of moles of KoH = 0.0011603 mol
Concentration of B in mol/dm^3 = (Number of moles of KoH) / Volume of B
Concentration of B in mol/dm^3 = 0.0011603 mol / (25.00 cm^3/1000 cm^3/dm^3)
Concentration of B in mol/dm^3 = 0.04641 mol/dm^3
Therefore, the concentration of B in mol/dm^3 is 0.04641 mol/dm^3.
C. Concentration of B in g/dm^3:
To find the concentration of B in g/dm^3, we need to convert the concentration from mol/dm^3 to g/dm^3.
Molar mass of KoH = 39 + 16 + 1 = 56 g/mol
Concentration of B in g/dm^3 = Concentration of B in mol/dm^3 x Molar mass of B
Concentration of B in g/dm^3 = 0.04641 mol/dm^3 x 56 g/mol
Concentration of B in g/dm^3 = 2.596 g/dm^3
Therefore, the concentration of B in g/dm^3 is 2.596 g/dm^3.