(D) For a $\frac{N}{10} \ NaOH$ solution,the concentration of $OH^-$ ions is $[OH^-] = 10^{-1} \ M$. Since $pOH = -\log[OH^-]$,we get $pOH = -\log(10^

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(D) For a $\frac{N}{10} \ NaOH$ solution,the concentration of $OH^-$ ions is $[OH^-] = 10^{-1} \ M$. Since $pOH = -\log[OH^-]$,we get $pOH = -\log(10^{-1}) = 1$. Using the relation $pH + pOH = 14$ at $25^{\circ}C$,we find $pH = 14 - 1 = 13$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)pH of strong Acids and strong Bases

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The $pH$ value of a $\frac{N}{10} \ NaOH$ solution is:

AIPMT 1996 All AIPMT

A
$10$
B
$11$
C
$12$
D
$13$

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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of strong Acids and strong Bases

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The $pH$ value of a $\frac{N}{10} \ NaOH$ solution is:

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What is the $pH$ of a $2.6 \times 10^{-8} \ M \ H^{+}$ ion solution?
$(\log 2.6 = 0.4150)$