14g of lithium hydroxide dissolves in 500cm^3 of water.in titration ,300m7^3 is used to neutralise sulphuric acid of 0,2mol.dm^-3 how do i calculate (a)concentation of a base and calculate for (c)volume of the acid and (c)which mass of Alkaline was used thanks in advance.
What is ,300m7^3?
Sorry My mistake it's 300m^3
How do i calculate a,b and c please help me
To calculate the concentration of the base and volume of acid used, you can use the following steps:
(a) Concentration of base:
Step 1: Convert the mass of lithium hydroxide (14g) to moles.
To do this, you need to know the molar mass of lithium hydroxide (LiOH), which can be calculated by adding the atomic masses of lithium (Li), oxygen (O), and hydrogen (H).
Molar mass of LiOH = atomic mass of Li + atomic mass of O + atomic mass of H
= (6.941 g/mol) + (15.999 g/mol) + (1.008 g/mol)
= 23.948 g/mol
Moles of LiOH = mass of LiOH / molar mass of LiOH
= 14g / 23.948 g/mol
≈ 0.584 mol
Step 2: Calculate the concentration of the base in mol/dm^3 (Molarity).
Given that the base dissolved in 500cm^3 (0.5dm^3) of water:
Concentration of base = moles of LiOH / volume of water
= 0.584 mol / 0.5 dm^3
= 1.168 mol/dm^3
So, the concentration of the base is approximately 1.168 mol/dm^3.
(b) Volume of the acid:
Given that 300cm^3 of the acid (0.3dm^3) was used, the volume of the acid is 0.3dm^3.
(c) Mass of alkaline used:
To calculate the mass of alkaline (LiOH) used, we need to use the stoichiometry of the balanced chemical equation between LiOH and H2SO4.
The balanced equation is: 2LiOH + H2SO4 → Li2SO4 + 2H2O
From the equation, we can see that 2 moles of LiOH react with 1 mole of H2SO4.
So, the moles of H2SO4 used can be calculated using the molarity and volume of the acid:
Moles of H2SO4 = Molarity x Volume in dm^3
= 0.2 mol/dm^3 x 0.3 dm^3
= 0.06 mol
Since 2 moles of LiOH react with 1 mole of H2SO4, the moles of LiOH used is also equal to 0.06 mol.
Finally, to calculate the mass of alkaline (LiOH), we can use the molar mass of LiOH and the moles of LiOH used:
Mass of LiOH = moles of LiOH x molar mass of LiOH
= 0.06 mol x 23.948 g/mol
= 1.437 g
Therefore, the mass of alkaline used is approximately 1.437 g.
Please note that all calculations are approximate and might vary slightly due to rounding.