A 35.0-mL sample of 0.20 M LiOH is titrated with 0.25 M HCl. What is the pH of the solution after 23.0 mL of HCl have been added to the base?
A)1.26
B)1.67
C)12.33
D)12.74
E)13.03

Question

Quizplus · Accepted Answer

The reaction between LiOH and HCl is a neutralization reaction, where LiOH (a strong base) reacts with HCl (a strong acid) to form LiCl (a salt) and water. The stoichiometry of the reaction is 1:1. First, calculate the moles of LiOH and HCl:Moles of LiOH = 0.035 L * 0.20 mol/L = 0.007 molMoles of HCl added = 0.023 L * 0.25 mol/L = 0.00575 molSince less moles of HCl are added than there are moles of LiOH, not all the LiOH will react. The remaining moles of LiOH:Remaining moles of LiOH = 0.007 mol - 0.00575 mol = 0.00125 molThe volume of the solution after adding HCl:Total volume = 35.0 mL + 23.0 mL = 58.0 mL = 0.058 LThe concentration of LiOH after the reaction:Concentration of LiOH = 0.00125 mol / 0.058 L = 0.02155 MSince LiOH is a strong base, its pH can be calculated using the formula:$$ \text{pOH} = -\log[\text{OH}^-] $$$$ \text{pH} = 14 - \text{pOH} $$$$ \text{pH} = 14 - (-\log[0.02155]) $$$$ \text{pH} \approx 12.33 $$Therefore, the pH of the solution after adding 23.0 mL of HCl is approximately 12.33.

A 350-ML Sample of 0