A 50.00-mL sample of 0.100 M KOH is titrated with 0.100 M HNO₃. Calculate the pH of the solution after the 52.00 mL of HNO₃ is added.

Question

Quizplus · Accepted Answer

The correct answer is D.The reaction between KOH and HNO3 is a neutralization reaction, which produces water and a salt. The balanced chemical equation for the reaction is:KOH + HNO3 -> H2O + KNO3The number of moles of KOH in the solution is:0.100 M * 0.050 L = 0.0050 molThe number of moles of HNO3 added is:0.100 M * 0.052 L = 0.0052 molSince the number of moles of HNO3 added is greater than the number of moles of KOH in the solution, the solution will be acidic. The pH of the solution can be calculated using the following equation:pH = -log[H+], where [H+] is the molar concentration of hydrogen ions in the solution.The molar concentration of hydrogen ions in the solution can be calculated using the following equation:[H+] = (moles of H+)/(total volume of solution)The total volume of the solution is:0.050 L + 0.052 L = 0.102 LThe number of moles of H+ in the solution is:0.0052 mol - 0.0050 mol = 0.0002 molTherefore, the molar concentration of hydrogen ions in the solution is:[H+] = 0.0002 mol / 0.102 L = 0.00196 MThe pH of the solution is:pH = -log(0.00196) = 2.71Therefore, the correct answer is D.

A 5000-ML Sample of 0