25 ml of a solution of barium hydroxide, on titration with 0.1 molar solution of HCl, gave a titre value of $35 m l$ . Molarity of barium hydroxide solution was:

0.07

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0.14

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0.28

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0.35

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Solution

The correct option is A 0.07
the titration reaction is:
$B a (O H)_{2} + 2 H C l ⟶ B a C l_{2}, + H_{2} O$
According to the reaction, the formula for a balanced equation is
$\frac{V o l u m e o f B a (O H)_{2} \times M o l a r i t y}{N o . o f m o l e c u l e s o f B a (O H)_{2}}$ $= \frac{V o l u m e o f H C l \times M o l a r i t y}{N o . o f m o l e c u l e s o f H C l}$
$= \frac{25 \times M o l a r i t y o f B a (O H)_{2}}{1}$ $= \frac{35 \times 0.1}{2}$
Therefore, molarity of barium hydroxide $= 0.07$

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