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Sulfamic acid, HSO3NH2 (molar mass = 97.1 g/mol), is a strong monoprotic acid that can be used to standardize a strong base: A 0.179-g sample of HSO3NH2 required 19.4 mL of an aqueous solution of KOH for a complete reaction. What is the molarity of the KOH solution?

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Respuesta :

Answer: The molarity of KOH solution is 0.093 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n = moles of solute

[tex]V_s[/tex] = volume of solution in ml

moles of [tex]HSO_3NH_2[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.179g}{97.1g/mol}=0.0018mol[/tex]

Now the balanced chemical equation will be :

[tex]HSO_3NH_2+KOH\rightarrow KSO_3NH_2+H_2O[/tex]

As 1 mole of [tex]HSO_3NH_2[/tex] requires 1 mole of KOH

Thus 0.0018 moles of [tex]HSO_3NH_2[/tex] requires =[tex]\frac{1}{1}\times 0.0018=0.0018[/tex] moles of KOH

Now put all the given values in the formula of molality, we get

[tex]Molarity=\frac{0.0018\times 1000}{19.4ml}[/tex]

[tex]Molarity=0.093M[/tex]

Therefore, the molarity of KOH solution is 0.093 M

Eduard22sly

The molarity of the KOH solution is 0.093 M

We'll begin by calculating the number of mole of HSO₃NH₂.

  • Mass of HSO₃NH₂ = 0.179 g
  • Molar mass of HSO₃NH₂ = 97.1 g/mol
  • Mole of HSO₃NH₂ =?

Mole = mass / molar mass

Mole of HSO₃NH₂ = 0.179 / 97.1

Mole of HSO₃NH₂ = 0.0018 mole

Next, we shall determine the number of mole of KOH required to react with 0.0018 mole of HSO₃NH₂.

HSO₃NH₂ + KOH —> KSO₃NH₂ + H₂O

From the balanced equation,

1 mole of HSO₃NH₂ reacted with 1 mole of KOH.

Therefore,

0.0018 mole of HSO₃NH₂ will also react with 0.0018 mole of KOH.

Finally, we shall determine the molarity of the KOH solution.

  • Mole of KOH = 0.0018 mole
  • Volume = 19.4 mL  = 19.4 / 1000 = 0.0194 L
  • Molarity of KOH =?

Molarity = mole / Volume

Molarity of KOH = 0.0018 / 0.0194

Molarity of KOH = 0.093 M

Thus, the molarity of the KOH solution is 0.093 M.

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