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. (Previous question ask: 10.00 mL of approximately 6 M sulfuric acid is transferred to a 100mL volumetric flask and diluted to the mark with distilled 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

Answer:- Actual molarity of the original sulfuric acid solution is 17.0M.

Solution:- Barium chloride reacts with sulfuric acid to make a precipitate of barium sulfate. The balanced equation is written as:

$BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)$

From this equation there is 1:1 mol ratio between barium sulfate and sulfuric acid. So, if excess of barium chloride is added to sulfuric acid then the moles of sulfuric acid would be equivalent to the moles of barium sulfate. Moles of barium sulfate could be calculated from the mass of it's dry precipitate.

Molar mass of barium sulfate is 233.4 grams per mol. The calculations for the moles of sulfuric acid are given below:

$0.397gBaSO_4(\frac{1molBaSO_4}{233.4gBaSO_4})(\frac{1moH_2SO_4}{1molBaSO_4})$

= $0.00170molH_2SO_4$

From given information, 10.00 mL of final acid solution were taken to react with excess of barium chloride. It means 0.00170 moles of sulfuric acid are present in 10.0 mL of final acid solution. We could calculate the actual molarity of the final solution from here as:

10.0 mL = 0.0100 L

$molarity=\frac{0.00170mol}{0.0100L}$

= 0.170M

Now we would use the dilution equation to calculate the actual molarity of the original sulfuric acid solution. The molarity equation is:

$M_1V_1=M_2V_2$

From given information, 10.0 mL of original acid solution were taken in a 100 mL flask and water was added up to the mark. It means the 10 fold dilution is done. 10 fold dilution means the molarity becomes one tenth of it's original value. Let's do the calculations in reverse way as we have calculated the molarity of the final solution.

let's say the molarity after first dilution is Y. the volume is taken as 10.0 mL. Final volume is 100 mL and the molarity is 0.170M. Let's plug in the values in the equation:

Y(10.0mL) = 0.170M(100mL)

$Y=\frac{0.170M*100mL}{10.0mL}Y = 1.70MLet's do the similar calculations to find out the actual molarity of the original acid solution. Let's say the molarity of the original acid solution is X. 10.0 mL of it were taken and diluted to 100 mL on adding water. The molarity is 1.70M as is calculated in the above step. Let's plug in the values in the molarity equation again to solve it for X as:X(10.0mL) = 1.70M(100mL)[tex]X=\frac{1.70M*100mL}{10.0mL}$

X = 17.0M

Hence, the actual molarity of sulfuric acid solution is 17.0M.