2/20/25, Mon An HBr solution has a concentration of 0.0150 M. What are the [H^+],[OH], pH, and pOH of this solution? A KOH solution has a concentration of 0 .0450 M What are the [OH],[H^+] pOH and oH of this

1. For the HBr solution:<br /> - The concentration of \(H^+\) ions in an acidic solution is equal to the concentration of the acid. Therefore, \([H^+] = 0.0150 \text{ M}\).<br /> - The concentration of \(OH^-\) ions in an acidic solution is calculated using the ion product constant for water, \(K_w = 1.0 \times 10^{-14}\). \([OH^-] = \frac{K_w}{[H^+]^2} = \frac{1.0 \times 10^{-14}}{(0.0150)^2} = 4.44 \times 10^{-12} \text{ M}\).<br /> - The pH of the solution is calculated using the formula \(pH = -\log[H^+]\). \(pH = -\log(0.0150) = 1.82\).<br /> - The pOH of the solution is calculated using the formula \(pOH = -\log[OH^-]\). \(pOH = -\log(4.44 \times 10^{-12}) = 11.35\).<br /><br />2. For the KOH solution:<br /> - The concentration of \(OH^-\) ions in a basic solution is equal to the concentration of the base. Therefore, \([OH^-] = 0.0450 \text{ M}\).<br /> - The concentration of \(H^+\) ions in a basic solution is calculated using the ion product constant for water, \(K_w = 1.0 \times 10^{-14}\). \([H^+] = \frac{K_w}{[OH^-]^2} = \frac{1.0 \times 10^{-14}}{(0.0450)^2} = 4.41 \times 10^{-12} \text{ M}\).<br /> - The pOH of the solution is calculated using the formula \(pOH = -\log[OH^-]\). \(pOH = -\log(0.0450) = 1.35\).<br /> - The pH of the solution is calculated using the formula \(pH = 14 - pOH\). \(pH = 14 - 1.35 = 12.65\).