Question

The pH of 0.001 M HCl solution is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Key Exam Tip:
For strong acids, $[\text{H}^+]$ = Molarity of acid. For strong bases, $[\text{OH}^-]$ = Molarity of base. Use $\text{pH} + \text{pOH} = 14$.

Answer 1

The Correct Option is C

Hydrochloric acid ($\text{HCl}$) is a strong acid, which means it dissociates completely in water. The dissociation reaction is: $\text{HCl (aq)} \rightarrow \text{H}^+ \text{(aq)} + \text{Cl}^- \text{(aq)}$ Since $\text{HCl}$ is a strong acid, the concentration of hydrogen ions ($\text{H}^+$) in the solution is equal to the initial concentration of $\text{HCl}$. Given that the molarity of the $\text{HCl}$ solution is $0.001 \text{ M}$. So, $[\text{H}^+] = 0.001 \text{ M}$. The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration: $\text{pH} = -\log_{10} [\text{H}^+]$ Substitute the concentration of $\text{H}^+$ into the formula: $\text{pH} = -\log_{10} (0.001)$ To calculate this, we can express $0.001$ in scientific notation: $0.001 = 1 \times 10^{-3}$ Now, substitute this into the pH equation: $\text{pH} = -\log_{10} (1 \times 10^{-3})$ Using the logarithm property $\log(a \times b) = \log a + \log b$: $\text{pH} = -(\log_{10} 1 + \log_{10} 10^{-3})$ We know that $\log_{10} 1 = 0$ and $\log_{10} 10^{-3} = -3$. $\text{pH} = -(0 + (-3))$ $\text{pH} = -(-3)$ $\text{pH} = 3$ Therefore, the pH of a $0.001 \text{ M HCl}$ solution is $3$.

Step 1: Recognize that $\text{HCl}$ is a strong acid. Strong acids completely dissociate in water, so the concentration of $\text{H}^+$ ions is equal to the initial concentration of the acid.

Step 2: Determine the hydrogen ion concentration. Given the concentration of $\text{HCl}$ is $0.001 \text{ M}$, the concentration of $\text{H}^+$ ions, $[\text{H}^+]$, is also $0.001 \text{ M}$.

Step 3: Write down the formula for pH. The pH is calculated using the formula: $\text{pH} = -\log_{10} [\text{H}^+]$.

Step 4: Substitute the hydrogen ion concentration into the pH formula. $\text{pH} = -\log_{10} (0.001 \text{ M})$

Step 5: Calculate the logarithm. Convert $0.001$ to scientific notation: $0.001 = 1 \times 10^{-3}$. $\text{pH} = -\log_{10} (1 \times 10^{-3})$ $\text{pH} = -(\log_{10} 1 + \log_{10} 10^{-3})$ $\text{pH} = -(0 + (-3))$ $\text{pH} = 3$ Final Answer: \(\boxed{\text{3}}\)