Q19.131CP - Chemistry: Molecular Nature Of Matter And Change

Question

A  $\mathbf{500}\mathbf{ }\mathbf{ml}$ solution consists of  $\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{ }\mathbf{mol}$ of solid NaOH  and  $\mathbf{0}\mathbf{.}\mathbf{130}\mathbf{ }\mathbf{mol}$ of hypochlorous acid $(\mathrm{HClO};K_a=3.0\times {10}^{-8})$  dissolved in water.

(a) Aside from water, what is the concentration of each species that is present?

(b) What is the  pH of the solution?

(c) What is the  pH after adding $\mathbf{0}\mathbf{.}\mathbf{005}\mathbf{ }\mathbf{mol}\mathbf{.}$ of  HCl to the flask?

Step-by-Step Solution

Answer

a) The concentration of ${\text{ClO}}^{\text{ - }} \text{is} \text{0.1 } \text{M } \text{and } \text{HClO} \text{is} \text{0.16} \text{M}$

b) The pH of the solution is $\mathbf{7}\mathbf{.}\mathbf{32}$.

C) After adding  $0.005 \mathrm{ml}$of  HCl to the flask then pH is  $7.25$.

Definition of pH

pH is a measure of hydrogen ion concentration, which indicates whether a solution is acidic or alkaline.

Concentration of each species

a)

Let us find the concentration of each species,

$$\begin{gathered} \text{V}=500\text{ml}=0.5 \text{L} \\ \text{n}\left(\text{NaOH}\right)=0.05 \text{mol} \\ \text{n}\left(\text{HClO}\right)=0.13 \text{mol} \end{gathered}$$ 

Let us write down the neutralization reaction taking place.

 $\text{HClO}+{\text{OH}}^{-}\leftrightharpoons {\text{ClO}}^{-}+{\text{H}}_2\text{O}$

From the balanced equation it is evident that  one mole of NaOH neutralizes one mole of HClO.

$0.05 \mathrm{mol} \mathrm{of} \mathrm{NaOH} \mathrm{neutralizes} 0.05 \mathrm{mol} \mathrm{of} \mathrm{HClO} \mathrm{to}\mathrm{form}0.05\mathrm{mol}\mathrm{of}{\mathrm{ClO}}^{-}$  

  $$\begin{gathered} \left[{\mathrm{ClO}}^{-}\right]=\frac{0.05\mathrm{mol}}{0.5\mathrm{L}} \\ =0.1\mathrm{M}. \\ \left[\mathrm{HClO}\right]=\frac{0.13\mathrm{mol}-0.05\mathrm{mol}}{0.5\mathrm{L}} \\ =0.16\mathrm{M}. \end{gathered}$$

Hence, the required concentration of ${\text{ClO}}^{\text{ - }} \text{is} \text{0.1} \text{M} \text{and} \text{HClO} \text{is} \text{0.16} \text{M}$. .

 

pH of the solution

b) 

Here we just have to apply Henderson-Hasselbalch equation:

$$\begin{gathered} \mathrm{pH}=\mathrm{pKa}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ \mathrm{pH}=-\log \left(3\times {10}^{-8}\right)+\log \frac{0.1\mathrm{M}}{0.16\mathrm{M}} \\ \mathrm{pH}=7.32. \end{gathered}$$ 

Therefore, the pH is $7.32$.

pH after adding 0.0050 mol of HCI

c) When $0.005 \mathrm{mol} \mathrm{of} \mathrm{HCl}$  is added the concentration of  H⁺  ions increase in the solution. According to the Le Chatlier’s principle, the back ward reaction will be favoured and concentration of   decrease. Taking these changes into account, we will use the Henderson-Hasselbalch equation to calculate the pH of the solution.

  $$\begin{gathered} \mathrm{pH}=-\log \left(3\times {10}^{-8}\right)+\log \frac{0.05 \text{mol}-0.005 \text{mol}}{0.13 \text{mol}-0.05 \text{mol}+0.005 \text{mol}} \\ =7.25. \end{gathered}$$

Therefore, after adding  $0.0050 \text{mol}$ of   to the flask then pH is $7.25$ .