Question

What approx volume of $0.40$\ $M$ $Ba(OH)_2$ must be added to $50.0$ $m\ell$ of $0.30$ $M$ $NaOH$ to get a solution in which the molarity of the $OH^-$ ions is $0.50$ $M$?

• A. $33\,m\ell$
• B. $66\,m\ell$
• C. $133\,m\ell$
• D. $100\,m\ell$

Answer 1

Answer: (A)

Molarity of $OH^- =$ $ \frac{Total \,moles\, of \,OH^{-}}{Total \,Vol.\, of \,solution}$
$0.50 = \frac{\left(2\,\times\,0.40\,\times\,V_{m\ell}\right)\,+\,0.30\,\times\,50}{V_{m\ell}\,+\,50}$
So $V = 33\,m\ell$