What is the molar concentration of h2so4 in a solution made by reacting 188.9 ml of 2.086 m h2so4 with 269.3 ml of 0.4607 m naoh?

0.7246 M

Explanation:

Considering:

Or,

Given :

For :

Molarity = 2.086 M

Volume = 188.9 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 188.9×10⁻³ L

Thus, moles of :

Moles of  = 0.39405 moles

For NaOH :

Molarity = 0.4607 M

Volume = 269.3 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 269.3×10⁻³ L

Thus, moles of NaOH :

Moles of NaOH  = 0.1241 moles

According to the given reaction:

1 moles of react with 2 moles of NaOH to form 1 mole of sodium sulfate.

Thus,

2 moles of NaOH react with 1 mole of

1 mole of NaOH react with 1/2 mole of

0.1241 moles of NaOH react with (1/2)×0.1241 mole of

Moles of that got reacted = 0.06205 moles

Unreacted moles = Total moles - Moles that got reacted = 0.39405 - 0.06205 moles = 0.332 moles

Total volume = 188.9×10⁻³ L + 269.3×10⁻³ L = 458.2×10⁻³ L

Concentration of :

Concentration of = 0.7246 M

the correct answer is answer a. a mixture

no other options make sense

hope this

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  - sprinkles

the answer is true.