Ionic EquilibriumHard
Question
What is the pH of 6.67 × 10−3 M aqueous solution of Al(OH)3 if its first dissociation is 100%, second dissociation is 50% and the third dissociation is negligible.
Options
A.2
B.12
C.11
D.3
Solution
$\left\lbrack OH^{-} \right\rbrack = 6.67 \times 10^{- 3} + \frac{6.67 \times 10^{- 3}}{2} + 0 \approx 10^{- 2}\text{ M}$
$\therefore P^{OH} = - \log\left( 10^{- 2} \right) = 2.0 \Rightarrow P^{H} = 12.0$
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
The solubility of AgCN in a buffer solution of pH = 3.0 is (Ksp of AgCN = 1.2 × 10−16; Ka of HCN = 4.8 × 10−10)...What is the pH of a neutral solution at 37°C, where Kw equals 2.5 × 10–14? (log 2 = 0.3)...Separate solutions of NaW, NaX, NaY and NaZ, each of concentrations 0.1 M has pH 7.0, 9.0, 10.0 and 11.0, respectively, ...What would be the pH of an ammonia solution if the pH of acetic acid solution of same strength is 3.2? The dissociation ...The active ingredient in aspirin is acetyl salicylic acidwith Ka = 4.0 × 10–9. The pH of the solution obtained by dissol...