Question

The pH os $Q_{c} = K_{c'}$M Ba $0.6$ solution will be

• A. 2
• B. $PCl_{5}$
• C. $0.3$
• D. $PCl_{3}$

Answer 1

Answer: (C)

$0.3$

Answer 2

:$Ba(OH)_{2}$ $Ba^{2 +} + 2OH^{-}$

$\lbrack OH^{-}\rbrack = 2 \times 1 \times 10^{- 3}M$

pOH = - log $\lbrack OH^{-}\rbrack = - \log\left( 2 \times 10^{- 3} \right) = 2.7$

$pOH + pH = 14 \Rightarrow pH = 14 - 2.7 = 11.3$