Question

Which of the following is/are true regarding a \(0.1 \mathrm{M}\) solution of \(\mathrm{Ba}(\mathrm{OH})_{2}\) ? (a) \(\left[\mathrm{OH}^{-}\right]=0.1 \mathrm{M}\) (b) \(\left[\mathrm{OH}^{-}\right]=\left[\mathrm{Ba}^{2+}\right]\) (c) \(\mathrm{pH}=13.30\) (d) \(\left[\mathrm{H}^{+}\right]=1 \times 10^{-7} \mathrm{M}\)

Short answer

Short answer: Among statements (a), (b), (c), and (d), only statement (c) is true. The pH of the 0.1 M Ba(OH)₂ solution is approximately 13.3.

Step-by-step solution

STEP 1: Write the dissociation equation for Barium hydroxide in water

Barium hydroxide dissociates in water as follows: \(\mathrm{Ba}(\mathrm{OH})_{2}(s) ->\mathrm{Ba}^{2+}(aq) + 2\mathrm{OH}^-(aq)\) This equation tells us that for every mole of \(\mathrm{Ba}(\mathrm{OH})_{2}\) that dissolves, there will be one mole of \(\mathrm{Ba}^{2+}\) and two moles of \(\mathrm{OH}^{-}\) ions released.

STEP 2: Find the concentration of \(\mathrm{OH}^{-}\) ions

The initial concentration of \(\mathrm{Ba}(\mathrm{OH})_{2}\) is given as \(0.1 \mathrm{M}\). Using the stoichiometry of the dissociation equation, we can find the concentration of \(\mathrm{OH}^{-}\) ions. For 1 mole of \(\mathrm{Ba}(\mathrm{OH})_{2}\) releasing 2 moles of \(\mathrm{OH}^{-}\) ions, the concentration of \(\mathrm{OH}^{-}\) ions will be: \([\mathrm{OH}^{-}] = 2 \times [\mathrm{Ba}(\mathrm{OH})_2] = 2 \times 0.1 \mathrm{M} = 0.2 \mathrm{M}\)

STEP 3: Compare our result to statement (a)

We can see that \([\mathrm{OH}^{-}] = 0.2 \mathrm{M}\) which is not equal to \(0.1 \mathrm{M}\). Therefore, statement (a) is false.

STEP 4: Compare our result to statement (b)

Since the concentration of \(\mathrm{OH}^{-}\) ions is \(0.2 \mathrm{M}\) and the concentration of \(\mathrm{Ba}^{2+}\) ions is \(0.1 \mathrm{M}\), we can see that \([\mathrm{OH}^{-}]\) is not equal to \([\mathrm{Ba}^{2+}]\). Therefore, statement (b) is also false.

STEP 5: Calculate the pOH and then the pH of the solution

The pOH of a solution is the negative base-10 logarithm of the hydroxide ion concentration. In this case, the pOH can be calculated as follows: \(\mathrm{pOH} = -\log_{10} (0.2)\) Now, to find the pH of the solution, we need to use the following relationship between pH and pOH: \(\mathrm{pH} + \mathrm{pOH} = 14\) \(\mathrm{pH} = 14 - \mathrm{pOH}\) Calculating the pH, we get: \(\mathrm{pH} = 14 - (-\log_{10}(0.2)) \approx 13.3\)

STEP 6: Compare our result to statement (c)

We found the pH of the solution to be approximately equal to 13.3. Therefore, statement (c) is true.

STEP 7: Compare our result to statement (d)

We know that the pH of the solution is approximately 13.3. The Hydronium ion concentration (\([\mathrm{H}^{+}]\)) can be found using the following equation: \([\mathrm{H}^{+}] = 10^{-\mathrm{pH}}\) \([\mathrm{H}^{+}] = 10^{-13.3}\) This value is not equal to \(1 \times 10^{-7} \mathrm{M}\). Therefore, statement (d) is false. In conclusion, out of the given statements, only statement (c) is true.