In a titration if 25.0cm3 of a 0.01moldm3 sodium hydroxide solution required 10.0cm3 of a tetraoxosulphate(vi)acid solution for complete neutralization
i) determine the moles of sodium hydroxide used in the reaction
ii)determine the moles of tetraoxosulphate(vi)acid used in the reaction
iii)determine the concentration of the tetraoxosulphate(vi)acid in
a.moldm3 b.gdm3
To determine the moles of sodium hydroxide used in the reaction, we can use the formula:
Moles = Volume (in dm^3) x Concentration
Given that the volume of sodium hydroxide used is 25.0 cm^3 (which is equivalent to 0.025 dm^3) and the concentration is 0.01 mol/dm^3, we can calculate:
Moles of sodium hydroxide = 0.025 dm^3 x 0.01 mol/dm^3 = 0.00025 mol
Therefore, the moles of sodium hydroxide used in the reaction is 0.00025 mol.
To determine the moles of tetraoxosulphate(vi)acid used in the reaction, we can use the mole ratio between sodium hydroxide and tetraoxosulphate(vi)acid. From the balanced equation of the reaction, we can see that the mole ratio is 1:1.
Thus, the moles of tetraoxosulphate(vi)acid used in the reaction is also 0.00025 mol.
To determine the concentration of the tetraoxosulphate(vi)acid in (a) moldm^3 and (b) gdm^3, we need to know the volume and mass of the tetraoxosulphate(vi)acid solution used.
If the volume of the tetraoxosulphate(vi)acid solution used is also 10.0 cm^3 (0.01 dm^3), then:
(a) Concentration of tetraoxosulphate(vi)acid in moldm^3 = Moles of tetraoxosulphate(vi)acid / Volume (in dm^3)
= 0.00025 mol / 0.01 dm^3
= 0.025 moldm^-3
(b) Concentration of tetraoxosulphate(vi)acid in gdm^3 = Concentration in moldm^3 x Molar mass of tetraoxosulphate(vi)acid
= 0.025 moldm^-3 x (Molar mass of tetraoxosulphate(vi)acid in g/mo)
Note: The molar mass of tetraoxosulphate(vi)acid (H2SO4) is approximately 98 g/mol.
= 0.025 moldm^-3 x 98 g/mol
= 2.45 gdm^-3
Therefore, the concentration of tetraoxosulphate(vi)acid in (a) moldm^3 is 0.025 moldm^-3 and (b) gdm^3 is 2.45 gdm^-3.