Question

pH of an aqueous solution containing NaOH is 12. The weight of NaOH present in 100mL of that solution in grams is:
A.0.4
B. 4.0
C. 0.04
D. 0.02

Answer 1

Hint From the pH find out the concentration of the hydronium ions and use that to find the moles of NaOH used. Now, from this data, use it to find the weight by multiplying it with the molar mass.

Complete step by step solution:
In order to answer the question, we need to know about pH. Now, there are 3 types of substances present which are acids, bases and salts. Acids contain ${{H}^{+}}$ ion and donate it, whereas bases contain $O{{H}^{-}}$ ions. That is why when an acid is mixed with water, these ions combine together and form ${{H}_{2}}O$ as a side product, along with the salt.
Now, let us talk about pH. pH is also named as the potential of hydrogen and it gives us an idea about how acidic a substance is. Now, pH is a scale that ranges from 0 to 14. 0 means that the substance is extremely acidic and 14 means that substance is extremely basic , or alkaline. 7 is the midpoint which indicates substance is neutral. Mathematically, $pH=-\log [{{H}^{+}}]$ . So, more the hydronium ion content in the acid, lesser is the pH and more acidic it is. Using the above formula of pH we can solve the question by finding out the value of $[{{H}^{+}}]$ . So, we have:

& pH=12 \\\ & \Rightarrow [{{H}^{+}}] ={{10}^{-12}}M \\\ \end{aligned}$$ Now, it is to be remembered that pH + pOH =14. So taking log and simplifying, we have $ [{{H}^{+}}] [O{{H}^{-}}] ={{10}^{-14}} $ . From this equation we will find out the $ [O{{H}^{-}}] $ , which is: $$\begin{aligned} & [O{{H}^{-}}] =\frac{{{10}^{-14}}}{[{{H}^{+}}] } \\\ & \Rightarrow [O{{H}^{-}}] =\frac{{{10}^{-14}}}{{{10}^{-12}}} \\\ & \Rightarrow [O{{H}^{-}}] ={{10}^{-2}}M \\\ \end{aligned}$$ Now, we will calculate the moles of NaOH using the molarity and volume. Using the data given in the question, we have: $$\begin{aligned} & mole{{s}_{NaOH}}=molarity\times volume \\\ & \Rightarrow mole{{s}_{NaOH}}={{10}^{-2}}\times 0.1 \\\ & \Rightarrow mole{{s}_{NaOH}}={{10}^{-3}}moles \\\ \end{aligned}$$ Now, we finally find the weight of NaOH required, that is: $$\begin{aligned} & weigh{{t}_{NaOH}}=moles\times molar\,mass \\\ & \Rightarrow weigh{{t}_{NaOH}}={{10}^{-3}}\times 40 \\\ & \Rightarrow weigh{{t}_{NaOH}}=0.04g \\\ \end{aligned}$$ **So, we get the weight as 0.04g which gives our correct answer as option C.** **NOTE:** It is to be noted that in the calculation of pH, the base of the logarithm is taken as 10. Whereas the base of ‘e’ is taken in the formula of natural logarithm.