Question

Calculate the ${\rm{pH}}$ of $\dfrac{{\rm{N}}}{{1000}}$ sodium hydroxide (NaOH) assuming complete ionization.
A.${\rm{pH}} = {\rm{9}}$
B. ${\rm{pH}} = 10$
C. ${\rm{pH}} = 11$
D. ${\rm{pH}} = 4$

Answer 1

Hint: We know that pH is the measure of concentration of hydrogen ion in a solution. The formula of pH is pH$ = - \log \left[ {{{\rm{H}}^ + }} \right]$. The formula indicates that, when the concentration of hydrogen ion is more in a solution, pH of the solution is less.

Complete step by step answer:
Here, the concentration of sodium hydroxide is given as $\dfrac{{\rm{N}}}{{1000}}$, that means, hydroxide concentration is $\dfrac{{\rm{N}}}{{1000}}$ or ${10^{ - 3}}$N.
$\left[ {{\rm{OH}}} \right] = {10^{ - 3}}$
Now, the calculation of pOH value of sodium hydroxide needs to be done. We know the formula of pOH, that is,
${\rm{pOH}} = - \log \left[ {{\rm{OH}}} \right]$ …………….…… (1)
Now, we have to put the value of hydroxide concentration in equation (1).
$ \Rightarrow {\rm{pOH}} = - \log \left[ {{{10}^{ - 3}}} \right] = - \left( { - 3} \right) = 3$
Therefore, pOH value is 3.
We know that the summation of pH and pOH is equal to 14. Now, we have to use this formula to calculate the pH value.
${\rm{pH}} + {\rm{pOH = 14}}$
$ \Rightarrow {\rm{pH}} = {\rm{14}} - {\rm{3}} = {\rm{11}}$

Therefore, pH of the sodium hydroxide solution is 11. Hence, the correct answer is option C.

Additional Information:
Always remember that an acid is a substance that can release protons in aqueous solution and a base is the substance that can accept protons in a solution. An acid after releasing proton is the conjugate base and the base after accepting proton forms the conjugate acid. These two are known as conjugate acid base pairs.

Note:
Strong acids are the acids that undergo complete dissociation such as HCl and strong bases are those bases that undergo complete dissociation, such as sodium hydroxide. Weak acids are those acids that undergo incomplete dissociation such as acetic acid and weak bases are those that undergo incomplete dissociation such as ammonia.