Answer to Question #309403 in General Chemistry for Sofia

Question

Determine the [OH−] , pH, and pOH of a solution with a [H+] of 6.2×10−13M at 25 °C.

[OH−]= M

pH=

pOH=

Determine the [H+] , pH, and pOH of a solution with an [OH−] of 3.9×10 −13 M at 25 °C.

[H+]= M

pH=

pOH=

Determine the [H+] , [OH−] , and pOH of a solution with a pH of 2.28 at 25 °C.

[H+]= M

[OH−]= M

pOH=

Determine the [H+], [OH−] , and pH of a solution with a pOH of 4.08 at 25 °C.

[H+]= M

[OH−]= M

pH=

Answer 1

pH + pOH = 14

pH = −log[H⁺]

pOH = −log[OH⁻]

[H⁺] = 10^-pH

[OH⁻] = 10^−pOH

(1) Determine the [OH⁻] , pH, and pOH of a solution with a [H⁺] of 6.2×10⁻¹³ M at 25 °C.

Solution (1):

pH = −log[H⁺] = −log(6.2×10⁻¹³) = 12.21

pOH = 14 − pH = 14 − 12.21 = 1.79

[OH⁻] = 10^−pOH = 10^−1.79 = 0.016 M = 1.6×10⁻² M

Therefore,

[OH⁻] = 1.6×10⁻² M

pH = 12.21

pOH = 1.79

(2) Determine the [H⁺] , pH, and pOH of a solution with an [OH⁻] of 3.9×10⁻¹³ M at 25 °C.

Solution (2):

pOH = −log[OH⁻] = −log(3.9×10⁻¹³) = 12.41

pH = 14 − pOH = 14 − 12.41 = 1.59

[H⁺] = 10^-pH = 10^-1.59 = 0.0257 = 2.57×10⁻² M

Therefore,

[H⁺] = 2.57×10⁻² M

pH = 1.59

pOH = 12.41

(3) Determine the [H⁺] , [OH⁻] , and pOH of a solution with a pH of 2.28 at 25 °C.

Solution (3):

[H⁺] = 10^-pH = 10^-2.28 = 5.25×10⁻³ M

pOH = 14 − pH = 14 − 2.28 = 11.72

[OH⁻] = 10^−pOH = 10^−11.72 = 1.9×10⁻¹² M

Therefore,

[H⁺] = 5.25×10⁻³ M

[OH⁻] = 1.9×10⁻¹² M

pOH = 11.72

(4) Determine the [H⁺], [OH⁻] , and pH of a solution with a pOH of 4.08 at 25 °C.

Solution (4):

[OH⁻] = 10^−pOH = 10^−4.08 = 8.3×10⁻⁵ M

pH = 14 − pOH = 14 − 4.08 = 9.92

[H⁺] = 10^-pH = 10^-9.92 = 1.2×10⁻¹⁰ M

Therefore,

[H⁺] = 1.2×10⁻¹⁰ M

[OH⁻] = 8.3×10⁻⁵ M

pH = 9.92