Question

Calculate the pH of 0.01 M strong dibasic acid.

• A. 5.5
• B. 2.5
• C. 2.0
• D. 1.7

Hint:

pH Tip: For a strong acid of molarity $C$ with basicity $n$ (number of displaceable $H^+$), $[H^+] = n \times C$. Always multiply the molarity by the basicity before taking the negative logarithm.

Answer 1

The Correct Option is D

Concept: Chemistry (Equilibrium) - pH Calculation of Strong Acids.

Step 1: Understand the nature of the given acid. A strong dibasic acid (like $H_2SO_4$ in its primary dissociation) is assumed to completely dissociate into its ions in an aqueous solution, yielding two moles of hydrogen ions ($H^+$ or $H_3O^+$) per mole of the acid.

Step 2: Determine the concentration of hydrogen ions. The molarity of the acid ($C$) is given as $0.01\text{ M}$. Since the acid is dibasic and fully dissociates, the concentration of hydrogen ions $[H_3O^+]$ will be twice the concentration of the acid. $[H_3O^+] = 2 \times C = 2 \times 0.01\text{ M} = 0.02\text{ M} = 2 \times 10^{-2}\text{ M}$.

Step 3: Set up the pH formula. The formula for pH is $pH = -\log_{10}[H_3O^+]$. Substitute the calculated hydrogen ion concentration: $pH = -\log_{10}(2 \times 10^{-2})$.

Step 4: Expand the logarithmic expression. Using the property of logarithms ($\log(AB) = \log A + \log B$): $pH = -(\log_{10} 2 + \log_{10} 10^{-2})$. Since $\log_{10} 10^{-2} = -2$, this becomes: $pH = -(\log_{10} 2 - 2) = 2 - \log_{10} 2$.

Step 5: Evaluate the final numerical value. The standard logarithmic value for $\log_{10} 2$ is approximately $0.3010$. Substitute this value into the equation: $pH = 2 - 0.3010 = 1.699$. Rounding to one decimal place to match the given options yields $1.7$. $$ \therefore \text{The pH of the solution is } 1.7. $$