Question

The pH of a mixture of $0.01M$ $\mathrm{HCl}$ and $0.1M$ ${\mathrm{CH}}_3\mathrm{COOH}$ is approximately

• A. $1$
• B. $2$
• C. $4$
• D. $7$

Answer 1

The correct option is A

$1$

Explanation for correct option

(A)$1$

Step1. Formula for pH

$\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

Step2. Calculation of concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$

• $\text{HCl}$ is a strong acid and when dissolved in water dissociates completely thus producing a large number of ${\mathrm{H}}^{+}$ ions
• Whereas $\text{C}{\text{H}}_3\mathrm{COOH}$ is a weak acid and it dissociates only to a small extent in the aqueous solution giving a small amount of ${\mathrm{H}}^{+}$ ions.

$$\begin{gathered} M_1{V}_1+M_2{V}_2=M_{\text{R}}\left(V_1+V_2\right) \\ 0.01\times 1000+0.1\times 1000=M_{\text{R}}\left(1000+1000\right) \\ {M}_{\text{R}}=5\times {10}^{-2} \end{gathered}$$

Step3. Calculation of pH

$$\begin{gathered} \mathrm{pH}=-\log \left[5\times {10}^{-2}\right]\left(\log \mathrm{mn}=\mathrm{logm}+\log \mathrm{n}\right) \\ =-\left[\log 5-2\log 10\right]\left({\mathrm{logm}}^{\mathrm{n}}=\mathrm{nlog}\mathrm{m}\right) \\ =2-\log 5 \\ =2-0.698 \\ =1.302 \\ =1 \end{gathered}$$

Hence, the value of pH is one.

Explanation for incorrect option

The pH of a mixture of $0.01M$ $\mathrm{HCl}$ and $0.1M$ ${\mathrm{CH}}_3\mathrm{COOH}$ is approximately $1$. Hence, options (B), (C), and (D) are incorrect.

Hence, option (A) is correct. The value of pH is $1$.