Question

A 50.0 mL sample of 0.040 M HCl is titrated with 0.040 M NaOH. What is the pH of the solution after 30.0 mL of NaOH have been added to the acid? I got Ph of 12, but the answer choices are: 3.85 2.96 2.00 2.49 1.81

a) 3.85
b) 2.96
c) 2.00
d) 2.49
e) 1.81

Answer 1

The correct pH of the solution after adding 30.0 mL of 0.040 M NaOH to 50.0 mL of 0.040 M HCl is 2.00, as the unreacted HCl is 0.010 M after the reaction.

Step-by-step explanation:

The question asks about the pH of a solution during a titration process. In this titration, we are adding 0.040 M NaOH to 0.040 M HCl. Initially, we have 50.0 mL of HCl which means we have 0.002 moles of HCl (50.0 mL × 0.040 mol/L). After adding 30.0 mL of 0.040 M NaOH, we thus have added 0.0012 moles of NaOH (30.0 mL × 0.040 mol/L).

Since HCl and NaOH react in a 1:1 molar ratio, the 0.0012 moles of NaOH will react with the same amount of HCl moles, leaving us with 0.002 - 0.0012 = 0.0008 moles of HCl. This remaining acid is in a total volume of 50.0 mL + 30.0 mL = 80.0 mL. To find the concentration of the remaining HCl solution, we divide the remaining moles by the total volume in liters: 0.0008 moles / 0.080 L = 0.010 M HCl.

The pH of a strong acid solution like HCl can be found by taking the negative logarithm (base 10) of its concentration. Hence, pH = -log(0.010) which is equal to 2.00. Therefore, the correct pH of the solution after adding 30.0 mL of NaOH is 2.00.