Constant‑boiling hcl can be used as a primary standard for acid–base titrations. a 50.00 ml sample of constant‑boiling hcl with a concentration of 0.1096 m was collected and titrated to an endpoint with 31.78 ml of a ba(oh)2 solution. what is the molarity of the ba(oh)2 solution

0.0862 M

Explanation:

Considering:

Or,

Given :

For :

Molarity = 0.1096 M

Volume = 50.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 50.00 × 10⁻³ L

Thus, moles of :

Moles of  = 0.00548 moles

According to the reaction,

2 moles of HCl react with 1 mole of

1 mole of HCl react with 1/2 mole of

0.00548 mole of HCl react with (1/2)*0.00548 mole of

Moles of = 0.00274 moles

Volume of = 31.78 mL = 31.78 × 10⁻³ L

Molarity = 0.00274 moles / 31.78 × 10⁻³ L  = 0.0862 M

yes it is balanced

the answer is c i believe