You visited us 1 times! Enjoying our articles? Unlock Full Access!

Degree of Dissociation

What is the f...

Question

What is the formula for determining $O H^{-}$ in a solution when given the concentration of protons in the solution?

[H_{3} O^{+}] [O H^{-}] = K w = 1 \times 10^{- 14}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[H_{3} O^{+}] [O H^{-}] = K w = 1 \times 10^{- 7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{[H_{3} O^{+}]}{[O H^{-}]} = K w = 1 \times 10^{- 14}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

K w = 1 \times 10^{- 14} \times [H_{3} O^{+}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

K_{w} \times [H_{3} O^{+}] [O H^{-}] = 1 \times 10^{- 14}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $[H_{3} O^{+}] [O H^{-}] = K w = 1 \times 10^{- 14}$
The formula for determining $O H^{-}$ in a solution (when given the concentration of the protons in the solution) is
$[H_{3} O^{+}] [O H^{-}] = K w = 1 \times 10^{- 14}$

Suggest Corrections

Similar questions

2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}, K_{w} = 1 \times 10^{- 14}

25^{\circ} C

K_{a}

K_{w}

2 H_{2} O ⇌ H_{3} O^{+} O H^{-}

10^{- 14}

25^{\circ} C

9.62 \times 10^{- 14}

60^{\circ} C

p H

60^{\circ} C

N H_{4}^{+} + 2 H_{2} O ⇌ N H_{4} O H + H_{3} O^{+}

K_{1} = 5.56 \times 10^{- 10}

H_{2} O + H_{2} O ⇌ H_{3} O^{+} + O H^{-}

K_{2} = 1 \times 10^{- 14}

K_{3}

N H_{4} O H ⇌ N H_{4}^{+} + O H^{-}

K_{w}

2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}

10^{- 14}

25^{o} C

9.62 \times 10^{- 14}

60^{o} C

p H

60^{o} C

\sqrt{9.60} = 3.10, log 3.10 = 0.49

K_{w}

2 H_{2} O ⇌ H_{3} O^{+} O H^{-}

10^{- 14}

25^{\circ}

9.62 \times 10^{- 14}

60^{\circ} C

p H

60^{\circ} C

Join BYJU'S Learning Program