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.
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
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