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 00450 M What are the [OH^-],[H^+] pOH, and pH of this solution? Calculate the [H+],[OH-] pH, and pOH of a solution.

To calculate the $[H^{+}]$, $[OH^{-}]$, pH, and pOH of a solution, we need to know the concentration of the solution and whether it is an acid or a base.<br /><br />For the HBr solution with a concentration of 0.0150 M:<br />Since HBr is a strong acid, it will dissociate completely in water to produce $H^{+}$ ions. Therefore, the concentration of $H^{+}$ ions will be equal to the concentration of the HBr solution, which is 0.0150 M.<br /><br />The concentration of $OH^{-}$ ions can be calculated using the ion product constant for water (Kw), which is $1.0 \times 10^{-14}$ at 25°C. The formula is:<br /><br />$[OH^{-}] = \frac{Kw}{[H^{+}]}$<br /><br />Substituting the values, we get:<br /><br />$[OH^{-}] = \frac{1.0 \times 10^{-14}}{0.0150} = 6.67 \times 10^{-12} M$<br /><br />The pH of the solution can be calculated using the formula:<br /><br />$pH = -\log[H^{+}]$<br /><br />Substituting the value of $[H^{+}]$, we get:<br /><br />$pH = -\log(0.0150) = 1.82$<br /><br />The pOH of the solution can be calculated using the formula:<br /><br />$pOH = -\log[OH^{-}]$<br /><br />Substituting the value of $[OH^{-}]$, we get:<br /><br />$pOH = -\log(6.67 \times 10^{-12}) = 11.18$<br /><br />For the KOH solution with a concentration of 0.0450 M:<br />Since KOH is a strong base, it will dissociate completely in water to produce $OH^{-}$ ions. Therefore, the concentration of $OH^{-}$ ions will be equal to the concentration of the KOH solution, which is 0.0450 M.<br /><br />The concentration of $H^{+}$ ions can be calculated using the ion product constant for water (Kw), which is $1.0 \times 10^{-14}$ at 25°C. The formula is:<br /><br />$[H^{+}] = \frac{Kw}{[OH^{-}]}$<br /><br />Substituting the values, we get:<br /><br />$[H^{+}] = \frac{1.0 \times 10^{-14}}{0.0450} = 2.22 \times 10^{-12} M$<br /><br />The pOH of the solution can be calculated using the formula:<br /><br />$pOH = -\log[OH^{-}]$<br /><br />Substituting the value of $[OH^{-}]$, we get:<br /><br />$pOH = -\log(0.0450) = 1.35$<br /><br />The pH of the solution can be calculated using the formula:<br /><br />$pH = 14 - pOH$<br /><br />Substituting the value of pOH, we get:<br /><br />$pH = 14 - 1.35 = 12.65$<br /><br />In summary, for the HBr solution with a concentration of 0.0150 M:<br />- $[H^{+}] = 0.0150 M$<br />- $[OH^{-}] = 6.67 \times 10^{-12} M$<br />- pH = 1.82<br />- pOH = 11.18<br /><br />For the KOH solution with a concentration of 0.0450 M:<br />- $[OH^{-}] = 0.0450 M$<br />- $[H^{+}] = 2.22 \times 10^{-12} M$<br />- pOH = 1.35<br />- pH = 12.65