An environmental chemist obtained a 200. mL sample of lake water believed to be contaminated with a single monoprotic strong acid. Titrating this sample with a 0.0050 M NaOH(aq) required 7.3 mL of the NaOH solution to reach the endpoint. What is the pH of the lake

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of NaOH used:
0.0050 M NaOH = 0.0050 mol/L
0.0050 mol/L x 0.0073 L = 3.65 x 10^-5 mol NaOH
Since the acid is monoprotic, the moles of NaOH used are equal to the moles of acid present in the sample.
Therefore, 3.65 x 10^-5 mol H+ are neutralized by the NaOH. 
We can calculate the concentration of H+ in the lake water using:
[H+] = (moles of H+)/volume of sample in liters
[H+] = (3.65 x 10^-5 mol)/(0.200 L) = 1.825 x 10^-4 M
Using the formula for strong acids, we know that pH = -log[H+]
pH = -log(1.825 x 10^-4) ≈ 3.74
Therefore, the best choice is B (3.74).

An Environmental Chemist Obtained a 200