Answer to Question #120069 in General Chemistry for aly

Question

What concentration of AlCl3 will produce a pH of 2.50 in 500ml of H2O

Answer 1

Solution:

AlCl₃=Al³⁺+3Cl^- - Aluminum chloride completely dissociates in an aqueous solution.

Al³⁺+H₂O=AlOH²⁺+H⁺ - only the first stage of hydrolysis proceeds significantly.

K_h=[AlOH²⁺][H⁺]/[Al³⁺] - constant of the first stage of Al³⁺ ion hydrolysis.

Multiplying the numerator and denominator of the hydrolysis constant expression by [OH^-], we get:

K_h=[AlOH²⁺][H⁺][OH^-]/([Al³⁺][OH^-])=K_w/K_bAlOH2+

K_w=[H⁺][OH^-]=1*10^-14 - Ionic product of water

K_bAlOH2+=[Al³⁺][OH^-]/[AlOH²⁺]=1*10^-9 - the constant of basicity of the AlOH²⁺ ion

For a dilute solution it will be fair:

[AlOH²⁺]=[H⁺]

and [Al³⁺]=C_AlCl3

Therefore

K_w/K_bAlOH2+=[H⁺]²/C_AlCl3

C_AlCl3=[H⁺]²K_bAlOH2+/K_w

pH=-lg[H⁺]

[H⁺]=10^-pH

So, C_AlCl3=10^-2pHK_bAlOH2+/K_w

C_AlCl3=10^-2*2.5*1*10^-9/1*10^-14=1(mol/L)

Answer:

Concentration of AlCl3 that will produce a pH of 2.50 in 500ml of H2O is 1mol/L.

And 0.5 mol of AlCl3 should be dissolved in 500ml of an aqueous solution.