calculate the moles of sulfuric acid that remained in the solution after reaction with the carbonate.
the tared mass of carbonate is 1.5179g
4 answers
You didn't post all of the problem.
i've calculated the moles of sulfuric acid present in the 50ml of standard acid solution to be 2.378x10^-1mol. the average volume of sodium hydroxide required for the back titration to be 31.03mL. and the moles of sodium hydroxide required for the back titration to be 2.951x10^-3. I'm not sure how to proceed
You've still posted portions although the last helps. My guess is that you added an excess of a standard H2SO4 soln to a CO3^= unknown and you're back titrating to find amount of excess H2SO4 added.
total mols H2SO4 added initially minus 1/2 the mols NaOH to back titrate gives you mols carbonate ion there originally in the unknown.
total mols H2SO4 added initially minus 1/2 the mols NaOH to back titrate gives you mols carbonate ion there originally in the unknown.
im guessing im missing the concentration of H2SO4 is 0.04755M and concentration for sulfuric acid is 0.09510M. In the flash i have 50.00ml of sulfuric acid and 25.00ml of impure sodium carbonate solution.
2 answers