Question

What is the ph of 0.450 m al(no3)3 [ka for al3+(aq) = 1.00x10-5]? express your answer to two decimal places?

Answer 1

Answer : The pH of the solution is, 2.67

Explanation :

The equilibrium chemical reaction is:

`Al^(3+)+H_2O\rightarrow Al(OH)^(2+)+H^+`

Initial conc. 0.450 0 0

At eqm. (0.450-x) x x

As we are given:

`K_a=1.00* 10^(-5)`

The expression for equilibrium constant is:

`K_a=((x)* (x))/((0.450-x))`

Now put all the given values in this expression, we get:

`1.00* 10^(-5)=((x)* (x))/((0.450-x))`

`x=0.00212M`

The concentration of
`H^+` = x = 0.00212 M

Now we have to calculate the pH of solution.

`pH=-\log [H^+]`

`pH=-\log (0.00212)`

`pH=2.67`

Therefore, the pH of the solution is, 2.67