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
What concentration of free CN– must be maintained in a solution that is 1.8 M-AgNO3 and 0.16 M-NaCl to prevent AgCl from...Ammonia forms the complex ion [Cu(NH3)4]2+ with copper ions in alkaline solutions but not in acidic solutions. What is t...Strong acids are generally used as standard solution in acid-base titrations because:...The degree of hydrolysis of a salt of weak acid and weak base in its 0.1 M solution is found to be 50%. If the molarity ...A solution of Na2CO3 is added drop by drop to litre of a solution containing 10-4 mole of Ba2+ and 10-5 mole of Ag+, if ...