25ml of a solution of na2c03 having a specific gravity of 1.25g ml required 32.9ml of a solution of hcl containing 109.5g of the acid per litre for complete neutralization. Calculate the volume of 0.84N Hes04 that will be completely neutralized by 125g of nac03 solution
The answer 470ml
Who helps
First calculate the molarity of Na2CO3 solution.
M HCl = 109.5g x (1 mol/36.5g) = 3.00M
mols HCl = M x L = 3.00 x 0.0329 = 98.0.09870.
mols Na2CO3 = 1/2 that (from the coefficients in the balanced equation) = 0.04935
Now the titration with H2SO4 (note: 0.84N = 0.42M)
Use density to convert 125 g Na2CO3 to volume. That's 100 mL
100 mL Na2CO3 x M Na2CO3 from above = mL H2SO4 x 0.42 and solve for mL H2SO4. 470 mL is correct.
To find the volume of 0.84N H2SO4 that will be completely neutralized by 125g of Na2CO3 solution, we can use the concept of stoichiometry and the given information.
First, let's calculate the molarity of the HCl solution. We know that 32.9 mL of a solution of HCl containing 109.5 g of the acid per litre is required for complete neutralization.
Molarity (M) = (Number of moles of solute) / (Volume of solution in liters)
Since we have the volume of the solution in milliliters, we need to convert it to liters:
Volume of solution (L) = 32.9 mL / 1000 = 0.0329 L
Now, we can calculate the molarity of the HCl solution using the formula:
Molarity of HCl = (Mass of solute / Molar mass of solute) / Volume of solution
Molar mass of HCl = 1(atomic mass of hydrogen) + 1(atomic mass of chlorine) = 1 + 35.5 = 36.5 g/mol
Molarity of HCl = (109.5 g / 36.5 g/mol) / 0.0329 L = 100 M
Now, let's determine the balanced equation for the reaction between Na2CO3 and H2SO4:
Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2
From the balanced equation, we can see that the stoichiometric ratio between Na2CO3 and H2SO4 is 1:1. This means that for every 1 mole of Na2CO3, we need 1 mole of H2SO4 for complete neutralization.
Next, calculate the number of moles of Na2CO3 in the given 125g of Na2CO3 solution using the formula:
Moles = (Mass of substance) / (Molar mass of substance)
Molar mass of Na2CO3 = 2(atomic mass of sodium) + 1(atomic mass of carbon) + 3(atomic mass of oxygen)
= 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 g/mol
Moles of Na2CO3 = 125 g / 106 g/mol = 1.179 mol
Since the stoichiometric ratio is 1:1 between Na2CO3 and H2SO4, we need 1.179 moles of H2SO4 for complete neutralization.
Now, let's calculate the volume of 0.84N H2SO4 that would contain 1.179 moles of H2SO4:
Molarity (N) = (Number of moles of solute) / (Volume of solution in liters)
0.84N = 1.179 mol / Volume of H2SO4 solution in liters
Solving for the volume:
Volume of H2SO4 solution = (1.179 mol) / (0.84N)
= 1.403 L = 1403 mL
Therefore, the volume of 0.84N H2SO4 that will be completely neutralized by 125g of Na2CO3 solution is 1403 mL or approximately 1400 mL.