Question

The $\text{pH}$ of a solution with $[\text{H}^+] = 3.2 \times 10^{-3} \text{ M}$ is closest to:

• A. 2.5
• B. 3.0
• C. 3.5
• D. 4.0

Hint:

To estimate $\text{pH}$ without a calculator for $[\text{H}^+] = C \times 10^{-N}$, remember $\text{pH} = N - \log(C)$. Since $\log(C)$ is positive, the $\text{pH}$ must be slightly less than $N$.

Answer 1

The Correct Option is A

We use the definition $\text{pH} = -\log[\text{H}^+]$.
Given $[\text{H}^+] = 3.2 \times 10^{-3} \text{ M}$.
$\text{pH} = -\log(3.2 \times 10^{-3})$.
$\text{pH} = -(\log(3.2) + \log(10^{-3}))$.
$\text{pH} = -(\log(3.2) - 3) = 3 - \log(3.2)$.
We use the approximation $\log(3.2) \approx 0.5$.
$\text{pH} \approx 3 - 0.5 = 2.5$.
The exact value is $3 - 0.505 = 2.495$, which is closest to 2.5.