Question2. Calculate $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{ClO}_{4}\right]$ and $[\mathrm{OH}]$ in an aqueous solution that is 0.150 M in $\mathrm{HClO}_{4}(\mathrm{aq})$. Is the solution acidic or basic?
3. Calculate, $\left[\mathrm{OH}^{-}\right],\left[\mathrm{K}^{+}\right]$and $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$in an aqueous solution that is 0.250 M in $\mathrm{KOH}(\mathrm{aq})$. Is the solution acidic or basic?
4. Compute $\left[\mathrm{Ca}^{2+}\right],\left[\mathrm{OH}^{-}\right]$and $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$for a solution that is prepared by dissolving 0.600 g of $\mathrm{Ca}(\mathrm{OH})_{2}$ in enough water to make $0.500 \mathrm{dm}^{3}$ of solution.

Studdy Solution
To find $[\mathrm{H}_3\mathrm{O}^+]$, use the ion product of water:
$$[\mathrm{H}_3\mathrm{O}^+] = \frac{K_w}{[\mathrm{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.0324}$$
$$[\mathrm{H}_3\mathrm{O}^+] \approx 3.09 \times 10^{-13} \, \mathrm{M}$$
The solutions are as follows:
1. For $\mathrm{HClO}_4$ solution: - $\left[\mathrm{H}_3\mathrm{O}^+\right] = 0.150 \, \mathrm{M}$ - $\left[\mathrm{ClO}_4^-\right] = 0.150 \, \mathrm{M}$ - $[\mathrm{OH}^-] \approx 6.67 \times 10^{-14} \, \mathrm{M}$ - The solution is acidic.
2. For $\mathrm{KOH}$ solution: - $[\mathrm{OH}^-] = 0.250 \, \mathrm{M}$ - $[\mathrm{K}^+] = 0.250 \, \mathrm{M}$ - $[\mathrm{H}_3\mathrm{O}^+] \approx 4.0 \times 10^{-14} \, \mathrm{M}$ - The solution is basic.
3. For $\mathrm{Ca(OH)}_2$ solution: - $[\mathrm{Ca}^{2+}] = 0.0162 \, \mathrm{M}$ - $[\mathrm{OH}^-] = 0.0324 \, \mathrm{M}$ - $[\mathrm{H}_3\mathrm{O}^+] \approx 3.09 \times 10^{-13} \, \mathrm{M}$ - The solution is basic.