50ml of 20.8% bacl2 and 100ml of 9.8% h2s04 solutions are mixed. Molarity of cl ions in the resulting solution wt of ba =137
I do not still understand step by step clearly
Thank a lot
To find the molarity of Cl ions in the resulting solution, we need to calculate the moles of Cl ions in both the BaCl2 and H2SO4 solutions separately, and then add them together.
Here's a step-by-step explanation:
Step 1: Calculate the moles of BaCl2:
Given: Volume of BaCl2 solution = 50 mL.
Concentration of BaCl2 = 20.8%.
Molar mass of BaCl2 = 137 g/mol (as given wt of Ba = 137).
First, convert the given volume from mL to liters:
Volume of BaCl2 solution = 50 mL = 0.05 L
Next, calculate the moles of BaCl2:
Moles of BaCl2 = (Concentration × Volume) / 1000
= (20.8 × 0.05) / 1000
= 0.0104 moles
Step 2: Calculate the moles of H2SO4:
Given: Volume of H2SO4 solution = 100 mL.
Concentration of H2SO4 = 9.8%.
Molar mass of H2SO4 = (2 × 1.008) + 32.07 + (4 × 16.00) = 98.09 g/mol
Convert the given volume from mL to liters:
Volume of H2SO4 solution = 100 mL = 0.1 L
Calculate the moles of H2SO4:
Moles of H2SO4 = (Concentration × Volume) / 1000
= (9.8 × 0.1) / 1000
= 0.0098 moles
Step 3: Calculate the moles of Cl ions:
In the BaCl2 solution, there is 1 mole of Cl ions per mole of BaCl2, since the formula of BaCl2 has a 1:2 ratio.
Therefore, the moles of Cl ions in the BaCl2 solution is equal to the moles of BaCl2, which we found in Step 1:
Moles of Cl ions from BaCl2 = 0.0104 moles
In the H2SO4 solution, there are 2 moles of HCl ions per mole of H2SO4, due to the balanced chemical equation for the reaction between H2SO4 and BaCl2.
Therefore, the moles of Cl ions from H2SO4 is equal to twice the moles of H2SO4, which we found in Step 2:
Moles of Cl ions from H2SO4 = 2 × 0.0098 moles = 0.0196 moles
Step 4: Add the moles of Cl ions from both solutions:
Total moles of Cl ions = moles of Cl ions from BaCl2 + moles of Cl ions from H2SO4
= 0.0104 moles + 0.0196 moles
= 0.03 moles
Step 5: Calculate the molarity of Cl ions in the solution:
Molarity (M) = Moles / Volume
= 0.03 moles / (50 mL + 100 mL) = 0.03 moles / 0.15 L
= 0.2 M
Therefore, the molarity of Cl ions in the resulting solution is 0.2 M.