Sulfamic acid, HSO₃NH₂ (molar mass = 97.1 g/mol) , is a strong monoprotic acid that can be used to standardize a strong base: HSO₃NH₂(aq) + KOH(aq) $\rightarrow$ KSO₂NH₂(aq) + H₂O(l)
A 0) 179-g sample of HSO₃NH₂ required 19.4 mL of an aqueous solution of KOH for complete reaction. What is the molarity of the KOH solution?

Question

Sulfamic acid, HSO₃NH₂ (molar mass = 97.1 g/mol) , is a strong monoprotic acid that can be used to standardize a strong base: HSO₃NH₂(aq) + KOH(aq)

\rightarrow

KSO₂NH₂(aq) + H₂O(l)
A 0) 179-g sample of HSO₃NH₂ required 19.4 mL of an aqueous solution of KOH for complete reaction. What is the molarity of the KOH solution?

Quizplus · Accepted Answer

To find the molarity of the KOH solution, first calculate the moles of sulfamic acid: 0.179 g / 97.1 g/mol = 0.001844 mol. Since the reaction is 1:1, moles of KOH = moles of sulfamic acid = 0.001844 mol. Molarity = moles/volume (L) = 0.001844 mol / 0.0194 L = 0.0950 M.

Sulfamic Acid, HSO3NH2 (Molar Mass = 97 →\rightarrow→ KSO2NH2(aq) + H2O(l) a 0)179-G Sample of HSO3NH2 Required 19

Sulfamic Acid, HSO₃NH₂ (Molar Mass = 97 $\rightarrow$ KSO₂NH₂(aq) + H₂O(l)
a 0)179-G Sample of HSO₃NH₂ Required 19