Question

The hydrolysis constant for the reaction $A{l}^{3+}+H_{2}O\rightleftarrows Al(OH{)}^{2+}+H^{+}$ is $1.7\times {10}^{-10}$ , What is $[H^{+}]$ in $0.1M\text{ AlC}{\text{l}}_3$ solution?
A. $4.00\times {10}^{-6}M$
B. $4.28\times {10}^{-6}M$
C. $4.12\times {10}^{-6}M$
D. None of these

Answer 1

Hint: An equilibrium constant for a hydrolysis reaction is called hydrolysis constant. In general hydrolysis constant for any reaction is equal to the ratio of the product of the concentration of hydronium ion and the concentration of the base to the concentration of the acid. Moreover, when a salt of an acid or base is dissolved in water and the interaction which occurs between the ions of the salt and water is called hydrolysis.

Complete answer:
According to the question: Given,
 $A{l}^{3+}+H_{2}O\rightleftarrows Al(OH{)}^{2+}+H^{+}$
The hydrolysis constant for the above reaction is $1.7\times {10}^{-10}$ . To calculate the concentration of $H^{+}$ in $0.1$ M in $AlC{l}_3$ .
Step 1: To calculate the initial concentration of each species:
The initial concentration of each species at time t=0.
 $[A{l}^{3+}]=0.1M$
 $[Al(OH{)}^{2+}]=0M$
 $[H^{+}]=0M$
Step 2: To calculate equilibrium concentration of each species:
The change in concentration of each species at time t=t.
 $[A{l}^{3+}]=0.1-xM$
 $[Al(OH){]}^{2+}=x$
 $[H^{+}]=x$
Step 3: To calculate the final concentration:
The hydrolysis constant is defined as an equilibrium constant for a hydrolysis reaction. The hydrolysis constant is denoted by the symbol $K_h$ . The hydrolysis constant for above reaction is
 $K_h={\displaystyle \frac{[H^{+}][Al{(OH)}^{2+}]}{[A{l}^{3+}]}}$
On substituting the values in the above equation, we get
 $1.7\times {10}^{-10}={\displaystyle \frac{x\times x}{0.1-x}}$
Since, the value of hydrolysis constant is very small, $0.1-x$ can be approximated of rounded off to $0.1$ , thereby ignoring the value of $x$
On solving the above equation, we get
 $x^2=1.7\times {10}^{-11}$
Or $x=4.12\times {10}^{-6}M$
Step 4: To substitute the obtained value of $x$ to determine the concentration of each species:
 $[A{l}^{3+}]=0.099M$
 $[Al(OH{)}^{2+}]=4.12\times {10}^{-6}$
 $[H^{+}]=4.12\times {10}^{-6}M$
The concentration of $H^{+}$ in $0.1MAlC{l}_3$ solution is $4.12\times {10}^{-6}M$ .
Therefore the correct answer is option C.

Note:
Aluminium chloride or aluminium trichloride with chemical formula $AlC{l}_3$ contains aluminium and chlorine atoms in $1:3$ ratio. Aluminium chloride participates in Friedel- crafts reaction as a Lewis- acid catalyst. Aluminium chloride can be manufactured on a large scale by the exothermic reaction of aluminium metal with either chlorine or hydrogen chloride.