(A) Given,$[OH^-] = 0.05 \ mol/L = 5 \times 10^{-2} \ mol/L$. $pOH = -\log[OH^-] = -\log(5 \times 10^{-2}) = 2 - \log(5) = 2 - 0.699 = 1.301$. Since $

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(A) Given,$[OH^-] = 0.05 \ mol/L = 5 \times 10^{-2} \ mol/L$. $pOH = -\log[OH^-] = -\log(5 \times 10^{-2}) = 2 - \log(5) = 2 - 0.699 = 1.301$. Since $pH + pOH = 14$ at $25^{\circ}C$,$pH = 14 - 1.301 = 12.699$. Since the $pH > 7$,the solution is basic.

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

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When a base is dissolved in water,it produces a hydroxide ion concentration of $0.05 \ mol/L$. Therefore,this solution will be.......

A
Basic
B
Acidic
C
Neutral
D
Both $A$ and $B$

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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of strong Acids and strong Bases

The highest $pH$ $(14)$ will be of which of the following?

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What is the $pH$ of a $NaOH$ solution with a concentration of $0.01 \ g \ L^{-1}$?

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Calculate the $pH$ of a $4 \times 10^{-3} \ M \ Y(OH)_2$ solution. Assume $100\%$ dissociation for the first step and $50\%$ dissociation for the second step.

Medium

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What is the $pH$ value of $\frac{N}{1000} \ KOH$ solution?

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When $10^{-8} \ mol$ of $HCl$ is dissolved in one litre of water,the $pH$ of the solution will be

Easy

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When a base is dissolved in water,it produces a hydroxide ion concentration of $0.05 \ mol/L$. Therefore,this solution will be.......

Similar Questions

The highest $pH$ $(14)$ will be of which of the following?

What is the $pH$ of a $NaOH$ solution with a concentration of $0.01 \ g \ L^{-1}$?

Calculate the $pH$ of a $4 \times 10^{-3} \ M \ Y(OH)_2$ solution. Assume $100\%$ dissociation for the first step and $50\%$ dissociation for the second step.

What is the $pH$ value of $\frac{N}{1000} \ KOH$ solution?

When $10^{-8} \ mol$ of $HCl$ is dissolved in one litre of water,the $pH$ of the solution will be

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