What is the molarity of chloride ion present in the solution that is formed when 238.50mL of 1.35M barium hydroxide reacts with 2146.5 mL of 0.3 M HCl?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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In this question, when barium hydroxide and hydrochloric acid react, we get water and barium chloride, with the reaction following the chemical equation: Ba(OH)2 + 2HCl --> BaCl2 + 2H2O.

Hydrochloric acid is the only reactant contributing the chloride ions. 2146.5 mL of 0.3 M HCl contains 2146.5*0.3/1000 = 0.643 mole of chloride ions.

The total volume of the solution that is formed when the reaction takes place is 238.5 + 2146.5 = 2385 mL or 2.385 L.

The molarity of a compound in a solution is the number of moles of the compound present in one liter of the solution.

Here, the molarity of the chloride ions is 0.643/2.385 = 0.269 M.

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lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

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the amount of moles = volume(ml)*molarity*10⁻³

the amount of HCl moles = 2146.5*0.3*10⁻³ ~ 0.64mol

the amount of BaCl₂moles = 238.5*1.35*10⁻³ ~ 0.32mol

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Ba(OH)₂+ 2HCl → BaCl₂ + 2H₂O

0.32mol  +  0.64 mol → 0.32mol + 0.64 mol

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BaCl₂→ 0.32mol 

BaCl₂→ Ba²⁺ + 2Cl

0.32mol  → 0.32mol + 0.64mol

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Cl → 0.64mol

2146.5ml + 238.5ml → 0.64mol

2385ml → 0.64mol

∴ 1000ml → 0.64*1000/2385 mol ~ 0.268mol

Hence Molarity = 0.268M

 

 

 

 

 

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