Q162CP - Chemistry: Molecular Nature Of Matter And Change

Question

Thiamine hydrochloride $\left({\text{C}}_{\text{12}}{\text{H}}_{\text{18}}{\text{ON}}_{\text{4}}{\text{SCl}}_{\text{2}}\right)$ is a water soluble form of thiamine (vitamin B1; ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}\mathbf{\text{ = 3.37}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{ - 7}}}$. How many grams of the hydrochloride must be dissolved in 10.00 mL of water to give a pH of 3.50?

Step-by-Step Solution

Answer

To produce a solution with a  pH = 3.50 a 1.00 g of ${\text{C}}_{\text{12}}{\text{H}}_{\text{18}}{\text{ON}}_{\text{4}}{\text{SCl}}_{\text{2}}$ (hydrochloride) is needed.

Step 1: Find the grams of the hydrochloride

${\text{C}}_{\text{12}}{\text{H}}_{\text{18}}{\text{ON}}_{\text{4}}\text{SCl}$ will dissociate in water.

${\text{C}}_{12}{\text{H}}_{18}{\text{ON}}_4{\text{SCl}}_2\to {\text{C}}_{12}{\text{H}}_{18}{\text{ON}}_4{\text{SCl}}^{+}+{\text{Cl}}^{-}$

Then the thiamine hydrochloride will react with water. It will transfer a proton to water.

${\text{C}}_{\text{12}}{\text{H}}_{\text{18}}{\text{ON}}_{\text{4}}{\text{SCl}}^{\text{ + }}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{C}}_{\text{12}}{\text{H}}_{\text{17}}{\text{ON}}_{\text{4}}{\text{SCl + H}}_{\text{3}}{\text{O}}^{\text{ + }}$

Now, solve for the $\left[{\text{H}}_3\text{O}\right]$  from the given pH.

$$\begin{gathered} \begin{array}{rcl}& & \text{pH = } \text{ - log}\left[{\text{H}}_{\text{3}}\text{O]}\right.\\ & & {\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{ = 10}}^{\text{ - pH}}{ \\ }^{} {\text{ = 10}}^{\text{3.50}}{ \\ }^{} \text{ = 3.16}\times {\text{10}}^{\text{ - 1}}.\end{array} \end{gathered}$$

Step 2: Solve for the concentration of the hydrochloride

To solve for the concentration of the hydrochloride, we need to manipulate its ICE table. Note that we want to obtain $\left[{\text{H}}_3{\text{O}}^{+}\right]=3.16\times {10}^{-4}\text{M.}$$\left[{\text{C}}_{12}{\text{H}}_{18}{\text{ON}}_4{\text{SCl}}^{+}\right]+{\text{H}}_2\text{O}\leftrightharpoons \left[{\text{H}}_3{\text{O}}^{+}\right]+ \left[{\text{C}}_{12}{\text{H}}_{17}{\text{ON}}_4\text{SCl}\right]$

Solve for 

$$\begin{gathered} {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{C}}_{\text{12}}{\text{H}}_{\text{17}}{\text{ON}}_{\text{4}}\text{SCl}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}{\left[{\text{C}}_{\text{12}}{\text{H}}_{\text{18}}{\text{ON}}_{\text{4}}\text{SCl + }\right]} \\ {\text{K}}_{\text{a}}\text{ = }\frac{\left(\text{3.16}\times {\text{10}}^{\text{ - 4}}\right)\left(\text{3.16}\times {\text{10}}^{\text{ - 4}}\right)}{{\text{X}}_{\text{i}}\text{ - 3.16}\times {\text{10}}^{\text{ - 4}}} \end{gathered}$$

Solve for ${\text{X}}_{\text{i}}.$

$\begin{array}{rcl}& & {\text{K}}_{\text{a}}\text{ = } \frac{\left(\text{3.16}\times {\text{10}}^{\text{ - 4}}\right)\left(\text{3.16}\times {\text{10}}^{\text{ - 4}}\right)}{{\text{X}}_{\text{i}}\text{ - 3.16}\times {\text{10}}^{\text{ - 4}}}{\text{X}}_{\text{i}}\text{ - 3.16}\times {\text{10}}^{\text{ - 4}}\\ & & {\text{ X}}_{\text{i}}\text{ = } \frac{\left(\text{3.16}\times {\text{10}}^{\text{ - 4}}\right)\left(\text{3.16}\times {\text{10}}^{\text{ - 4}}\right)}{{\text{K}}_{\text{a}}}\\ & & \text{ = } \frac{\left(\text{3.16}\times {\text{10}}^{\text{ - 4}}\right)\left(\text{3.16}\times {\text{10}}^{\text{ - 4}}\right)}{{\text{K}}_{\text{a}}}\text{ + 3.16}\times {\text{10}}^{\text{ - 4}}\\ & & \text{ = } \frac{\left(\text{3.16}\times {\text{10}}^{\text{ - 4}}\right)\left(\text{3.16}\times {\text{10}}^{\text{ - 4}}\right)}{\text{3.37}\times {\text{10}}^{\text{ - 7}}}\text{ + 3.16}\times {\text{10}}^{\text{ - 4}}\\ & & \text{ = } \text{0.297 M.}\end{array}$

Therefore, we need ${\text{0.297M of C}}_{\text{12}}{\text{H}}_{\text{18}}{\text{ON}}_{\text{4}}{\text{SCl}}_{\text{2}}$ . Now, solve for the needed thiamide hydrochloride to produce 10.00 mL solution.

Step 3: Solve for the amount of moles

First, solve for the amount of moles needed.

$\begin{array}{rcl}& & \text{ M} \text{ = } \frac{\text{mol}}{\text{L}}\\ & & \text{ mol} \text{ = } \text{M}\times \text{L}\\ & & \text{ = } \text{0.297M}\times \text{10.00mL}\times \frac{\text{1L}}{\text{1000mL}}\text{mol }\\ & =& 2.97\times {10}^{-3}\mathrm{mol}\\ & & \text{mass = } \text{2.97}\times {\text{10}}^{\text{ - 3}}\text{mol×}\frac{337.3 \mathrm{g}}{\mathrm{mol}}\\ & & \text{=1.00 g}\end{array}$

Therefore, 1.00 g of ${\text{C}}_{\text{12}}{\text{H}}_{\text{18}}{\text{ON}}_{\text{4}}{\text{SCl}}_{\text{2}}$ is needed to produce a 10.00 mLsolution with a    pH = 3.50. 

Hence, 1.00 g of  ${\text{C}}_{\text{12}}{\text{H}}_{\text{18}}{\text{ON}}_{\text{4}}{\text{SCl}}_{\text{2}}$( hydrochloride).