What approx volume of 0.40 M Ba(OH)2 must be added to 50.0 mℓ of 0.30 M NaOH to get a solution in which the molarity of the OH- ions is 0.50 M?

Question

What approx volume of 0.40 M Ba(OH)2 must be added to 50.0 mℓ of 0.30 M NaOH to get a solution in which the molarity of the OH- ions is 0.50 M? (A) 33 mℓ (B) 66 mℓ (C) 133 mℓ (D) 100 mℓ

Tardigrade Admin · Accepted Answer

Molarity of OH- = (Total moles of OH-/Total Vol. of solution) 0.50 = ((2 × 0.40 × Vmℓ) + 0.30 × 50/Vmℓ + 50) So V = 33 mℓ

Q. What approx volume of $0.40$ \ $M$ $B a (O H)_{2}$ must be added to $50.0$ $m ℓ$ of $0.30$ $M$ $N a O H$ to get a solution in which the molarity of the $O H^{-}$ ions is $0.50$ $M$ ?

$33 m ℓ$

$66 m ℓ$

$133 m ℓ$

$100 m ℓ$

Molarity of $O H^{-} =$ $\frac{T o t a l m o l es o f O H ^{-}}{T o t a l V o l . o f so l u t i o n}$
$0.50 = \frac{( 2 \times 0.40 \times V _{m ℓ} ) + 0.30 \times 50}{V _{m ℓ} + 50}$
So $V = 33 m ℓ$

Q. What approx volume of 0.40\ M Ba(OH)2​ must be added to 50.0 mℓ of 0.30 M NaOH to get a solution in which the molarity of the OH− ions is 0.50 M?

33mℓ

66mℓ

133mℓ

100mℓ