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Question: \[NaOH\] is a strong base. What will be the pH of \[5.0\times {{10}^{-2}}\] M \[NaOH\] solution? \[\...

NaOHNaOH is a strong base. What will be the pH of 5.0×10−25.0\times {{10}^{-2}} M NaOHNaOH solution? [log 2=0.3].\left[ \text{log }2=0.3 \right].
A.14.0014.00
B.13.7013.70
C.13.0013.00
D.12.7012.70

Explanation

Solution

We know that the buffer is, what Henderson equation is. The solutions which resist the change in pH are known as buffer solutions. Buffer solution contains weak acid/ base and their salts. An aqueous solution of weak acid and its conjugate base or weak base and its conjugate acid is known as buffer.

Complete answer:
As we know that the pH depends upon the hydrogen and hydroxide ion concentrations. pH increases on increasing the concentration of hydroxide ions. The pH decreases on increasing hydrogen ion concentration. The effect of addition of strong acid or base in the buffer is explained by the Le-Chatelier’s principle. The buffer solution is a pH resistance solution. The pH of the buffer solution does not alter the addition of strong acid or base.
Now let us understand what a buffer solution is. Buffer solution is a solution which consists of a mixture that contains weak acid and its conjugate base or weak base and its conjugate acid. Buffer will help to maintain the pH of a solution. Buffers resist pH change on the addition of an acidic or basic component. It has the ability to neutralize small amounts of added acid or base while maintaining the pH of the solution for stable reaction. Also given that 5.0×10−2M  NaOH≡[OH−]=5×10−2M5.0\times {{10}^{-2}}M~~NaOH\equiv \left[ O{{H}^{-}} \right]=5\times {{10}^{-2}}M
As we know that the [H+][OH−]=1×10−14⇒[H+]=1×10−14[OH−]\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]=1\times {{10}^{-14}}\Rightarrow \left[ {{H}^{+}} \right]=\dfrac{1\times {{10}^{-14}}}{\left[ O{{H}^{-}} \right]}
On substituting we get;
⇒[H+]=1×10−145×10−2=2×10−13\Rightarrow \left[ {{H}^{+}} \right]=\dfrac{1\times {{10}^{-14}}}{5\times {{10}^{-2}}}=2\times {{10}^{-13}}
Thus, pH=−log⁡[H+]=−log⁡(2×10−13)=12.69≈12.70pH=-\log \left[ {{H}^{+}} \right]=-\log \left( 2\times {{10}^{-13}} \right)=12.69\approx 12.70
∴pH=−log⁡[H+]=12.70.\therefore pH=-\log \left[ {{H}^{+}} \right]=12.70.
Therefore, the correct answer is option D.

Note:
Remember that in many reactions and solutions, buffers are used as a means of keeping pH at a nearly constant value in a variety of chemical reactions. The pH change is very less when a small amount of strong acid or base is added to the buffer solution.