Q22.57CP - Chemistry: Molecular Nature Of Matter And Change

Question

Tetraphosphorus decaoxide $\left({\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\right)$ is made from phosphate rock and used as a drying agent in the laboratory. 

(a) Write a balanced equation for its reaction with water. 

(b) What is the pH of a solution formed from the addition of 8.5 g of  $\mathbf{\left(}{\mathbf{\text{P}}}_{\mathbf{\text{4}}}{\mathbf{\text{O}}}_{\mathbf{\text{10}}}\mathbf{\right)}$ in sufficient water to form $\mathbf{\text{0.750}}$ L? (See Table  18.5, p. 803 , for additional information.)

Step-by-Step Solution

Answer

(a) The required balanced equation is  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{(s) + 6H}}_{\text{2}}\text{O(l)}\to {\text{4H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{(l)}$.

(b) The  of the solution is  $1.515$.

Step 1: Definition of Elements in nature and Industry

 Chemically, elements are substances that cannot be broken down into simpler things. Hydrogen (H), with an atomic number of 1, is one of nature's elements. This element gave origin to all others and makes up 75% of the mass of the cosmos. Carbon (C) is a six-atomic element.

Step 2: Write a balanced equation

(a)  

The tetraphosphorus decaoxide $\left({\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\right)$ is used as a drying agent. Phosphoric acid is generated in the presence of water.

Thus, the balanced equation is  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{(s) + 6H}}_{\text{2}}\text{O(l)}\to {\text{4H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{(l)}$.

Step 3: What is the pH of a solution formed

(b)

 

 At first, the mole number of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ is calculated as,

 $$\begin{gathered} n\left(P_4{O}_{10}\right)=\frac{m}{M_R} \\ =\frac{8.50 g}{283.8890 g/mol} \\ n\left(P_4{O}_{10}\right)=0.02994 mol \end{gathered}$$

Now, based on the reaction equation, from 1 mol of  $P_4{O}_{10},4 mol$ of  ${\text{H}}_3{\text{PO}}_4$ is formed. Thus,

 $$\begin{gathered} n\left(H_{3}P{O}_4\right)=4\times n\left(P_4{O}_{10}\right) \\ n\left(H_{3}P{O}_4\right)=0.11976 mol \end{gathered}$$

If the solution volume is  $0.750 \text{L}$, the molarity of the solution would be

  $$\begin{gathered} \text{ Molarity }=\frac{n}{k} \\ \text{Molarity }=\frac{0.11976 \text{mol}}{0.750 \text{L}} \\ \text{Molarity }=0.15968\text{M} \end{gathered}$$

Because phosphoric acid is a weak acid that only partially dissociates in water to generate ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}$ , the pH of the solution cannot be estimated directly.

For phosphoric acid, there are three acid dissociation constants  $\left({\text{k}}_{\text{a}}\right)$, one for each of the protons lost. Based on the Table  ,

 ${\text{k}}_{\text{a,1}}{\text{ > > k}}_{\text{a,2}}{\text{ > > k}}_{\text{a,3}}$

As a result, only the first proton loss is relevant, while the other two can be ignored (since the  and  are so little).

The first acid dissociation constant is written as follows:

 

 $$\begin{gathered} k_{a,1}=\frac{\left[H^{+}\right]\times \left[H_{2}P{O}_4^{-}\right]}{\left[H_{3}P{O}_4\right]} \\ 7.2·{10}^{-3}=\frac{\left[H^{+}\right]·\left[H_{2}P{O}_4^{-}\right]}{\left[H_{3}P{O}_4\right]} \end{gathered}$$

 

Since the amount of ${\text{H}}^{+}$ and  ${\text{H}}_2{\text{PO}}_4^{-}$ produced is the same (based on the reaction equation), assign its concentration as  $x$:

 $\left[{\text{H}}^{\text{ + }}\right]\text{ = }\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}\right]\text{ = x}$

 

Since the starting concentration of  is computed, when   of phosphoric acid is lost during dissociation, the phosphoric acid concentration at equilibrium is:

 ${\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}_{\text{eq}}\text{ = (0.15968 - x)M}$

After that, the entire equation can be rewritten as

 $$\begin{gathered} 7.2\times {10}^{-3}=\frac{x\times x}{0.15968-x} \\ 7.2\times {10}^{-3}\times (0.15968-x)=x^2 \\ {x}^2-7.2\times {10}^{-3}\times (0.15968-x)=0 \\ {x}^2+7.2\times {10}^{-3}x-1.1497\times {10}^{-3}=0 \end{gathered}$$

How to solve the quadratic equation for  $x$,

 $$\begin{gathered} x=\frac{-7.2·{10}^{-3}\pm \sqrt{{\left(-7.2·{10}^{-3}\right)}^2-4·1·\left(-1.1497·{10}^{-3}\right)}}{2·1} \\ x=\frac{-7.2·{10}^{-3}\pm \sqrt{4.65064·{10}^{-3}}}{2·1} \\ x=\frac{-7.2·{10}^{-3}\pm 0.068196}{2·1} \\ \mathbf{x}=\mathbf{0}.\mathbf{0305}\text{M} \end{gathered}$$

As a result, the concentration of ${\text{H}}^{\text{ + }}$ ion in the solution is  $\text{0.0305mM}$. Finally, the solution's $\text{pH}$  value is

 $$\begin{gathered} pH=-\log \left[H^{+}\right] \\ pH=-\log [0.0305] \\ pH=1.515 \end{gathered}$$

 

Therefore, the  of the solution is  $1.515$.