Problem 44 - Chemistry

Question

Calculate the \(\left[\mathrm{H}^{+}\right]\) of each of the following solutions at \(25^{\circ} \mathrm{C}\). Identify each solution as neutral, acidic, or basic. a. \(\left[\mathrm{OH}^{-}\right]=1.5 \mathrm{M}\) b. \(\left[\mathrm{OH}^{-}\right]=3.6 \times 10^{-15} \mathrm{M}\) c. \(\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-7} M\) d. \(\left[\mathrm{OH}^{-}\right]=7.3 \times 10^{-4} M\)

Step-by-Step Solution

Answer

The concentrations of \(\mathrm{[H^+]}\) for the given solutions are: a. \(\mathrm{[H^+]} = 6.67 \times 10^{-15} M\) (basic) b. \(\mathrm{[H^+]} = 2.78 \times 10^{0} M\) (basic) c. \(\mathrm{[H^+]} = 1.0 \times 10^{-7} M\) (neutral) d. \(\mathrm{[H^+]} = 1.37 \times 10^{-12} M\) (basic)

Step 1: Calculate \(\mathrm{[H^+]}\) for Solution a

To find the \(\mathrm{[H^+]}\) for solution a, where \(\mathrm{[OH^-]} = 1.5 \mathrm{M}\), use the equation: \([\mathrm{H^+}] = \frac{K_w}{[\mathrm{OH^-}]}\) \([\mathrm{H^+}] = \frac{1.0 \times 10^{-14}}{1.5}\) \([\mathrm{H^+}] = 6.67 \times 10^{-15}\) As \(\mathrm{[H^+]} < \mathrm{[OH^-]}\), solution a is basic.

Step 2: Calculate \(\mathrm{[H^+]}\) for Solution b

To calculate the \(\mathrm{[H^+]}\) for solution b, where \(\mathrm{[OH^-]} = 3.6 \times 10^{-15} \mathrm{M}\), use the equation: \([\mathrm{H^+}] = \frac{K_w}{[\mathrm{OH^-}]}\) \([\mathrm{H^+}] = \frac{1.0 \times 10^{-14}}{3.6 \times 10^{-15}}\) \([\mathrm{H^+}] = 2.78 \times 10^{-15}\) As \(\mathrm{[H^+]} < \mathrm{[OH^-]}\), solution b is basic.

Step 3: Calculate \(\mathrm{[H^+]}\) for Solution c

To calculate the \(\mathrm{[H^+]}\) for solution c, where \(\mathrm{[OH^-]} = 1.0 \times 10^{-7} \mathrm{M}\), use the equation: \([\mathrm{H^+}] = \frac{K_w}{[\mathrm{OH^-}]}\) \([\mathrm{H^+}] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-7}}\) \([\mathrm{H^+}] = 1.0 \times 10^{-7}\) As \(\mathrm{[H^+]} = \mathrm{[OH^-]}\), solution c is neutral.

Step 4: Calculate \(\mathrm{[H^+]}\) for Solution d

To calculate the \(\mathrm{[H^+]}\) for solution d, where \(\mathrm{[OH^-]} = 7.3 \times 10^{-4} \mathrm{M}\), use the equation: \([\mathrm{H^+}] = \frac{K_w}{[\mathrm{OH^-}]}\) \([\mathrm{H^+}] = \frac{1.0 \times 10^{-14}}{7.3 \times 10^{-4}}\) \([\mathrm{H^+}] = 1.37 \times 10^{-12}\) As \(\mathrm{[H^+]} < \mathrm{[OH^-]}\), solution d is basic. To summarize, the nature of the solutions are: a - basic, b - basic, c - neutral, and d - basic.