Number of moles of hydroxide ions in 0.3L of 0.005M solution of Ba (OH...
0.3 L = 0.3 L of 0.005 M Ba(OH)2
0.3 x 0.005 = 0.0015 moles of Ba(OH)2
It will dissociate into 0.0015 moles of Ba and 2 x 0.0015 = 0.003 moles of OH ions.
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Number of moles of hydroxide ions in 0.3L of 0.005M solution of Ba (OH...
Understanding the Problem
To calculate the number of moles of hydroxide ions in a solution of barium hydroxide (Ba(OH)2), we need to consider both the concentration of the solution and the dissociation of Ba(OH)2 in water.
Dissociation of Barium Hydroxide
- Barium hydroxide, Ba(OH)2, dissociates in water as follows:
- Ba(OH)2 → Ba²⁺ + 2OH⁻
- This means that for every one mole of Ba(OH)2 that dissolves, it produces two moles of hydroxide ions (OH⁻).
Calculating Moles of Ba(OH)2
- We are given:
- Volume of the solution = 0.3 L
- Molarity (M) = 0.005 M
- The number of moles of Ba(OH)2 can be calculated using the formula:
- Moles = Molarity × Volume
- Moles of Ba(OH)2 = 0.005 M × 0.3 L = 0.0015 moles
Calculating Moles of Hydroxide Ions
- Since each mole of Ba(OH)2 produces 2 moles of OH⁻, we multiply the moles of Ba(OH)2 by 2:
- Moles of OH⁻ = 0.0015 moles of Ba(OH)2 × 2 = 0.003 moles
Conclusion
Thus, the total number of moles of hydroxide ions in the 0.3 L of 0.005 M solution of Ba(OH)2 is 0.003 moles, confirming that the correct answer is option 'B'.