Question

Sulfamic acid is a primary standard that can be used to standardize$\mathbf{NaOH}\mathbf{.}$

${}^{\mathbf{+}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{NSO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}\mathbf{OH}\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}{\mathbf{NSO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$Sulfamic acidFM 97.094What is the molarity of a sodium hydroxide solution if$\mathbf{34}\mathbf{.}\mathbf{26}\mathbf{}\mathbf{mL}$react with$\mathbf{0}\mathbf{.}\mathbf{3337}\mathbf{}\mathbf{g}$of sulfamic acid?

Short answer

The molarity of NaOH is${\mathrm{M}}_{\mathrm{NaOH}}=0.1001\mathrm{MNaOH}$

Step-by-step solution

Define the molarity.

The total number of moles of solute per litre of solution is known as a solution's molarity. Because, unlike mass, the volume of a system fluctuates with changes in physical conditions, the molality of a solution is affected by changes in physical parameters of the system such as pressure and temperature. Molarity is represented by the letter M, which stands for molar.

The molarity of a solution is defined as one gram of solute dissolved in one litre of solution.

Step 2:Calculate the molarity of NaOH

$$\begin{gathered} \mathrm{volume}\mathrm{of}\mathrm{NaOH} \\ \mathrm{Solution}=34.26\mathrm{mL}=0.03426\mathrm{L} \\ \mathrm{mass}\mathrm{of}{\mathrm{H}}_3{\mathrm{NSO}}_3=0.3337\mathrm{g} \\ {\mathrm{FM}}_3\mathrm{of}{\mathrm{H}}_3{\mathrm{NSO}}_3=97.094\mathrm{g}/\mathrm{mol} \\ \mathrm{By}\mathrm{using}\mathrm{the}\mathrm{mole}\mathrm{concept}\mathrm{and}\mathrm{molarity}\mathrm{formula}\mathrm{to}\mathrm{calculate}\mathrm{the}\mathrm{molarity}\mathrm{of}\mathrm{NaOH} \\ {\mathrm{M}}_{\mathrm{NaOH}}=\frac{0.333{\mathrm{gH}}_3{\mathrm{NSO}}_3\left({\displaystyle \frac{1{\mathrm{molH}}_3{\mathrm{NSO}}_3}{97.094{\mathrm{gH}}_3{\mathrm{NSO}}_3}}\right)\left({\displaystyle \frac{1\mathrm{molNaOH}}{1{\mathrm{molH}}_3{\mathrm{NSO}}_3}}\right)}{0.03426\mathrm{L}} \\ =0.1001\frac{\mathrm{molNaOH}}{\mathrm{L}} \\ =0.1001\mathrm{MNaOH} \\ \mathrm{Thus},\mathrm{The}\mathrm{molarity}\mathrm{of}\mathrm{NaOH}\mathrm{is}{\mathrm{M}}_{\mathrm{NAOH}}=0.1001\mathrm{MNaOH} \end{gathered}$$