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Question: The number of moles of hydroxide \((O{{H}^{-}})\) ion in 0.3 litre of 0.005 M solution of \(Ba(O{{H}...

The number of moles of hydroxide (OH)(O{{H}^{-}}) ion in 0.3 litre of 0.005 M solution of Ba(OH2)Ba(O{{H}_{2}})is:
(A)- 0.0050
(B)- 0.0030
(C)- 0.0015
(D)- 0.0075

Explanation

Solution

Molarity represented by M is the amount of a substance in a certain volume of solution. Molarity also defined as the molar concentration of a solution is defined as the number of moles of solute present in one litre of a solution. The molarity is given by-
Molarity(M)=Number of moles of soluteVolume of solution (in litres)Molarity(M)=\frac{\text{Number of moles of solute}}{\text{Volume of solution (in litres)}}

Complete step by step solution:
-Let us first write the dissociation of barium hydroxide molecule-
Ba(OH)2Ba2++2OHBa{{(OH)}_{2}}\rightleftharpoons B{{a}^{2+}}+2O{{H}^{-}}
It is clear from the above reaction that one mole of barium hydroxide produces 2 moles of hydroxide ions.
-According to question,
Molarity of Ba(OH)2Ba{{(OH)}_{2}} = 0.005 M
Volume of solution = 0.3 litre
Therefore, the number of moles = Molarity×Volume of solution (in litre) Molarity\times \text{Volume of solution (in litre) }
= 0.005 ×0.3=0.0015moles=1.5×103moles\text{= 0}\text{.005 }\times 0.3=0.0015moles=1.5\times {{10}^{-3}}moles
-Each boron hydroxide gives = 22 OHO{{H}^{-}} ions
Therefore the number of hydroxide ions produced = 2×1.5×103=0.003mol2\times 1.5\times {{10}^{-3}}=0.003mol

Hence, the correct answer is option B.

Note: You should not get confused between molarity and molality. The ratio of moles of a solute to the volume of solution in a litre is known as molarity. On the other hand, the ratio of the moles of a solute to the kilograms of a solvent is known as the molality. Molarity is represented by ‘M’ whereas molality is represented by ‘M’. Molarity depends on the volume of the whole solution whereas molality depends on the mass of the solvent. Molarity has a unit as moles/litre whereas molality has the unit as moles/kg. However, molarity and molality shares a common relationship which is given by the equation given below-
m=M×1000(d×1000)M(M)m=\frac{M\times 1000}{(d\times 1000)-M(M')}
Where mm = Molality
MM = Molarity
DD = Density
MM'= Molar mass of the solute