Question

in the laboratory a student combines 27.2 ml of a .338 M barium bromide solution with 17.7 ml of a .566 M barium acetate solution, what is the final concentration of barium cation?

Answer 1

Q160904

In the laboratory, a student combines 27.2 ml of a .338 M barium bromide solution with 17.7 ml of a .566 M barium acetate solution, what is the final concentration of barium cation?

Solution:

Barium bromide and barium acetate both are soluble in an aqueous medium.

We are given the concentration and volume of BaBr₂ and Ba(C₂H₃O₂)₂ solution.

Using this we can find the moles of BaBr₂ and Ba(C₂H₃O₂)₂.

From moles of BaBr₂ and Ba(C₂H₃O₂)₂ we can find the moles of Ba⁺² present in both the

compounds.

Total moles of Ba⁺² after mixing will be equal to the sum of moles of Ba⁺² present in both compounds.

Once we got the total moles of Ba⁺², we can use the total moles of Ba⁺² and total volume in the molarity formula and find the final concentration of Barium cation.

Step 1: To find moles of BaBr₂

We are given 27.2 ml of a .338 M barium bromide solution.

27.2 mL in 'L' = $27.2\space mL * \frac{1\space L}{1000\space mL} = 0.0272\space L ;$

$molarity = \frac{ moles \space of\space BaBr_2}{volume \space of \space BaBr_2 \space in\space 'L' }$

$0.338\space M = \frac{ moles \space of\space BaBr_2}{0.0272\space L}$

multiply both the side by 0.0272L we have

$0.338\space mol/L * 0.0272\space L = \frac{ moles \space of\space BaBr_2}{0.0272\space L} * 0.0272\space L$ ; ( since M = mol/L)

0.009194 mol = moles of BaBr₂

Step 2: To find the moles of Ba(C₂H₃O₂)₂.

We are given 17.7 ml of a .566 M barium acetate solution.

17.7 mL in 'L' = $17.7\space mL * \frac{1\space L}{1000\space mL} = 0.0177\space L ;$

$molarity = \frac{ moles \space of\space Ba(C_2H_3O_2)_2}{volume \space of \space Ba(C_2H_3O_2)_2 \space in\space 'L' }$

$0.566\space M = \frac{ moles \space of\space Ba(C_2H_3O_2)_2}{0.0177\space L}$

multiply both the side by 0.0177L we have

$0.566\space mol/L * 0.0177\space L = \frac{ moles \space of\space Ba(C_2H_3O_2)_2}{0.0177\space L} * 0.0177\space L$ ; ( since M = mol/L)

0.01002 mol = moles of Ba(C₂H₃O₂)₂.

Step 3: To find the total moles of Ba⁺² ion.

The dissociation of BaBr₂ in the solution is given as

BaBr₂ ===> Ba⁺² (aq) + 2 Br^- (aq) ;

The mole to mole ratio of BaBr₂ and Ba⁺² in the reaction is 1 :1.

so moles of Ba⁺² obtained from BaBr₂ will be the same, that is 0.009194 mol Ba⁺²

The dissociation of Ba(C₂H₃O₂)₂. is given as

Ba(C₂H₃O₂)₂. ===> Ba⁺² (aq) + 2C₂H₃O₂^-1 (aq)

The mole to mole ratio of Ba(C₂H₃O₂)₂. and Ba⁺² in the reaction is 1 :1.

so moles of Ba⁺² obtained from Ba(C₂H₃O₂)₂. will be the same, that is 0.01002 mol Ba⁺²

Total moles of Ba⁺² = 0.009194 mol Ba⁺² + 0.01002 mol Ba⁺²

= 0.019214 mol Ba⁺²

Step 4: To find the final concentration of Ba⁺² after mixing.

Total volume of the solution = 27.2mL + 17.7 mL = 44.9 mL.

Final volume in 'L' = $44.9\space mL * \frac{1\space L}{1000\space mL} = 0.0449\space L ;$

Plug 0.019214 mol Ba⁺² and 0.0449 L in the molarity formula and find the final concentration of

Ba⁺² cation.

$molarity = \frac{ moles \space of\space Ba^{+2}}{volume \space of \space final \space solution \space in\space 'L' }$

$molarity = \frac{ 0.019214 \space mol \space of\space Ba^{+2}}{0.0449L} = 0.428mol/L ;$

Hence the final concentration of Ba⁺² cation in the solution is 0.428 M or 0.428 mol/L.