Question

what is the molarity of a sodium hydroxide solution if 28.7 ml of this solution reacts exactly with 22.30 ml of 0.253 m sulfuric acid?

Answer 1

`\Huge \boxed{\boxed{\textbf{0.03932 M (4 s.f)}}}`

Step-by-step explanation:

To find the molarity of the sodium hydroxide solution, we first need to consider the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4):

`\LARGE \boxed{\tt{H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O}}`

From the balanced equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide. Given that 22.30 mL of 0.253 M sulfuric acid reacts exactly with 28.7 mL of the sodium hydroxide solution, we can calculate the moles of sulfuric acid:

`\large \boxed{\tt{Moles\, of\, H_2SO_4 = 22.30 \, mL * (0.253 \, mol)/(1000 \, mL) = 0.0056419 \, mol}}`

Since 1 mole of
`\tt{H_2SO_4}` reacts with 2 moles of NaOH, we can find the moles of NaOH:

`\large \boxed{\tt{Moles\, of\, NaOH = 2 * 0.0056419 \, mol = 0.0112838 \, mol}}`

Now, we can calculate the molarity of the sodium hydroxide solution:

`\large \boxed{\tt{Molarity\, of\, NaOH = (0.0112838 \, mol)/(28.7 \, mL) * (1000 \, mL)/(1 \, L) = 0.03931637631 \, M}}`

So, the molarity of the sodium hydroxide solution is approximately 0.393 M.

#BTH1

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