What is the $[O H^{-}]$ in the final solution prepared by mixing $20.0 m L$ of $0.050 M$ $H C l$ with $30.0 m L$ of $0.10 M B a (O H)_{2}$ ?

0.12 M

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0.10 M

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0.40 M

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0.0050 M

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Solution

The correct option is A $0.10 M$
$B a (O H)_{2} + 2 H C l \to B a C l_{2} + 2 H_{2} O$

2 m mol of HCl neutralize 1 m mole of $B a (O H)_{2}$

1 m mol of HCl neutralize 0.5 m mol of $B a (O H)_{2}$

$B a (O H)_{2}$ left = 3 - 0.5 m mol = 2.5 m mol

$[B a (O H)_{2}] = \frac{2.5}{50} M = 0.05 M$

or $[O H]^{-} = 2 \times 0.05 = 0.1 M$

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What is the [OH−] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0mL of 0.10MBa(OH)2?

The correct option is A 0.10 MBa(OH)2+2HCl→BaCl2+2H2O2 m mol of HCl neutralize 1 m mole of Ba(OH)21 m mol of HCl neutralize 0.5 m mol of Ba(OH)2Ba(OH)2 left = 3 - 0.5 m mol = 2.5 m mol [Ba(OH)2]=2.550M=0.05Mor [OH]−=2×0.05=0.1M

What is the $[O H^{-}]$ in the final solution prepared by mixing $20.0 m L$ of $0.050 M$ $H C l$ with $30.0 m L$ of $0.10 M B a (O H)_{2}$ ?