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:

$K_a=1.00\times 10^{-5}$

The expression for equilibrium constant is:

$K_a=\frac{(x)\times (x)}{(0.450-x)}$

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

$1.00\times 10^{-5}=\frac{(x)\times (x)}{(0.450-x)}$

$x=0.00212M$

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