Question

10.00 mL of the final acid solution is reacted with excess barium chloride to produce a precipitate of barium sulfate (Fw: 233.4 g/mol). The dry solid weighs 0.397 g. Use this mass and the dilution volumes to calculate the actual molarity of the sulfuric acid in the initial solution. *(In a previous question 10.00 mL of approx. 6 M sulfuric acid is transferred to a 100mL volumetric flask and diluted to the mark with distillied water and mixed. Then 10.00 mL of this solution was further diluted to 100 mL. The molarity of the final solution was 0.06 M).

Answer 1

BCE
\[H _{2}SO _{4}(aq) + BaCl _{2}(aq) \rightarrow BaSO _{4}(s) + 2HCl\]

So far, I did:
\[0.397gBaSO _{4} \times \frac{ 1 molBaSO _{4} }{ 233.4 g BaSO _{4} }\times \frac{ 1 mole H _{2} SO _{4} }{ 1moleBaSO _{4} }\times \frac{ 98.074gH _{2}SO _{4} }{ 1 moleH _{2}SO _{4} }=0.1668g\]
Which gave me the mass of the sulfuric acid.

Then, I did...
\[0.1668gH _{2}SO _{4}\times \frac{ 1 mole H _{2}SO _{4} }{ 98.078g H _{2}SO _{4}}= 0.001701 moles H _{2}SO _{4} \]
Finally, I did:
\[\frac{ 0.001701moles H _{2}SO _{4} }{ 0.01L }= 0.17 M H _{2}SO _{4}\]

Is this correct? If not, where did I go wrong? Thank you in advance for your input!

Answer 2

2HCl(aq)**

Answer 3

@Jhannybean If you get a minute, can you look at the work above and if you know how to solve it let me know if I did that correctly? If not, can you give me some pointers as to where I went wrong? If not, I understand. Thanks in advance if you can!

Answer 4

What you calculated is the concentration of sulfuric acid in the final acid solution. If I've understood the question, they want the concentration for the initial acid solution. You just need to use M1V1=M2V2 to go backwards for each dilution previously done.

Otherwise looks good!