Question

Determine the [OH −

], pH, and pOH of a solution with a [H +
]of 4.9×10 −13
M at 25 ∘
C. [OH −
]= pOH= Determine the [H ∗
]⋅pH, and pOH of a solution with an [OH −
]of 8.8×10 −13
M at 25 ∘
C. [H ∗
]= pH= Determine the ∣H +
∣,∣OH −
], and pOH of a solution with a pH of 3.72 at 25 ∘
C. [H +
]= [OH −
]= pOH= Determine the [H ∗
],[OH −
], and pH of a solution with a pOH of 9.45 at 25 ∘
C.

Answer 1

Determine the [OH^-], pH, and pOH of a solution with a [H^+] of 4.9×10^-13 M at 25 °C.

Given:

`\displaystyle\sf [H^(+)] = 4.9 * 10^(-13) \, M`

To find [OH^-], we can use the equation for the ion product of water:

`\displaystyle\sf [H^(+)][OH^(-)] = 1.0 * 10^(-14)`

Since we know [H^+], we can rearrange the equation to solve for [OH^-]:

`\displaystyle\sf [OH^(-)] = (1.0 * 10^(-14))/([H^(+)])`

`\displaystyle\sf [OH^(-)] = (1.0 * 10^(-14))/(4.9 * 10^(-13))`

`\displaystyle\sf [OH^(-)] \approx 2.04 * 10^(-2) \, M`

To find the pH, we can use the equation:

`\displaystyle\sf pH = -\log[H^(+)]`

`\displaystyle\sf pH = -\log(4.9 * 10^(-13))`

`\displaystyle\sf pH \approx 12.31`

To find the pOH, we can use the equation:

`\displaystyle\sf pOH = -\log[OH^(-)]`

`\displaystyle\sf pOH = -\log(2.04 * 10^(-2))`

`\displaystyle\sf pOH \approx 1.69`

Therefore, in the given solution, [OH^-] ≈ 2.04×10^-2 M, pH ≈ 12.31, and pOH ≈ 1.69.

-------------------

Determine the [H^+], pH, and pOH of a solution with an [OH^-] of 8.8×10^-13 M at 25 °C.

Given:

`\displaystyle\sf [OH^(-)] = 8.8 * 10^(-13) \, M`

Using the ion product of water equation, we can find [H^+]:

`\displaystyle\sf [H^(+)][OH^(-)] = 1.0 * 10^(-14)`

`\displaystyle\sf [H^(+)] = (1.0 * 10^(-14))/([OH^(-)])`

`\displaystyle\sf [H^(+)] = (1.0 * 10^(-14))/(8.8 * 10^(-13))`

`\displaystyle\sf [H^(+)] \approx 1.14 * 10^(-2) \, M`

To find the pH, we can use the equation:

`\displaystyle\sf pH = -\log[H^(+)]`

`\displaystyle\sf pH = -\log(1.14 * 10^(-2))`

`\displaystyle\sf pH \approx 1.94`

To find the pOH, we can use the equation:

`\displaystyle\sf pOH = -\log[OH^(-)]`

`\displaystyle\sf pOH = -\log(8.8 * 10^(-13))`

`\displaystyle\sf pOH \approx 12.06`

Therefore, in the given solution, [H^+] ≈ 1.14×10^-2 M, pH ≈ 1.94, and pOH ≈ 12.06.

-------------------

Determine the |H^+|, |OH^-|, and pOH of a solution with a pH of 3.72 at 25 °C.

Given:

`\displaystyle\sf pH = 3.72`

To find [H^+], we can use the equation:

`\displaystyle\sf [H^(+)] = 10^(-pH)`

`\displaystyle\sf [H^(+)] = 10^(-3.72)`

`\displaystyle\sf [H^(+)] \approx 2.2 * 10^(-4) \, M`

Since water is neutral, [H^+][OH^-] = 1.0×10^-14. Therefore, we can find [OH^-]:

`\displaystyle\sf [OH^(-)] = (1.0 * 10^(-14))/([H^(+)])`

`\displaystyle\sf [OH^(-)] = (1.0 * 10^(-14))/(2.2 * 10^(-4))`

`\displaystyle\sf [OH^(-)] \approx 4.55 * 10^(-11) \, M`

To find the pOH, we can use the equation:

`\displaystyle\sf pOH = -\log[OH^(-)]`

`\displaystyle\sf pOH = -\log(4.55 * 10^(-11))`

`\displaystyle\sf pOH \approx 10.34`

Therefore, in the given solution, |H^+| ≈ 2.2×10^(-4) M, |OH^-| ≈ 4.55×10^(-11) M, and pOH ≈ 10.34.

-------------------

Determine the [H^+], [OH^-], and pH of a solution with a pOH of 9.45 at 25 °C.

Given:

`\displaystyle\sf pOH = 9.45`

To find [OH^-], we can use the equation:

`\displaystyle\sf [OH^(-)] = 10^(-pOH)`

`\displaystyle\sf [OH^(-)] = 10^(-9.45)`

`\displaystyle\sf [OH^(-)] \approx 3.52 * 10^(-10) \, M`

Since water is neutral, [H^+][OH^-] = 1.0×10^-14. Therefore, we can find [H^+]:

`\displaystyle\sf [H^(+)] = (1.0 * 10^(-14))/([OH^(-)])`

`\displaystyle\sf [H^(+)] = (1.0 * 10^(-14))/(3.52 * 10^(-10))`

`\displaystyle\sf [H^(+)] \approx 2.84 * 10^(-5) \, M`

To find the pH, we can use the equation:

`\displaystyle\sf pH = -\log[H^(+)]`

`\displaystyle\sf pH = -\log(2.84 * 10^(-5))`

`\displaystyle\sf pH \approx 4.55`

Therefore, in the given solution, [H^+] ≈ 2.84×10^-5 M, [OH^-] ≈ 3.52×10^-10 M, and pH ≈ 4.55.