(A) Given the concentration of hydronium ions: $[H_3O^+] = 1 \times 10^{-10} \, M$. First,calculate the $pH$ of the solution: $pH = -\log[H_3O^+] = -\

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(A) Given the concentration of hydronium ions: $[H_3O^+] = 1 \times 10^{-10} \, M$. First,calculate the $pH$ of the solution: $pH = -\log[H_3O^+] = -\log[10^{-10}] = 10$. At $25 \, ^\circ C$,the relationship between $pH$ and $pOH$ is: $pH + pOH = 14$. Substituting the value of $pH$: $10 + pOH = 14$. Therefore,$pOH = 14 - 10 = 4$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Water hydrolysis and pH scale

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Calculate the $pOH$ of a solution at $25 \, ^\circ C$ that contains $1 \times 10^{-10} \, M$ of hydronium ions,i.e.,$H_3O^{+}$.

AIPMT 2007 All AIPMT

A
$4$
B
$9$
C
$1$
D
$7$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Water hydrolysis and pH scale

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Calculate the $pOH$ of a solution at $25 \, ^\circ C$ that contains $1 \times 10^{-10} \, M$ of hydronium ions,i.e.,$H_3O^{+}$.

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