A 50.00-mL sample of 0.100 M KOH is titrated with 0.202 M HNO₃.Calculate the pH of the solution after 52.00 mL of HNO₃ is added.

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Quizplus · Accepted Answer

Before the titration, we have a solution of KOH which is a strong base. We can assume that all of the KOH will ionize to form OH- ions in solution. Thus, we can find the concentration of OH- ions in the solution using the formula: [OH-] = Molarity of KOH = 0.100 M At the start of the titration, we have added 0 mL of HNO₃, so the concentration of OH- ions is still 0.100 M. During the titration, HNO₃ will react with the OH- ions to form water, using the equation: HNO₃ + OH- → H2O + NO3- We can use stoichiometry to find the number of moles of HNO₃ that have reacted with the OH- ions. n(HNO₃) = Molarity of HNO₃ x Volume of HNO₃ added n(HNO₃) = 0.202 M x 0.052 L = 0.0105 mol From the balanced equation above, we can see that 1 mole of HNO₃ reacts with 1 mole of OH-. Therefore, the number of moles of OH- ions that have reacted is also 0.0105 mol. The remaining concentration of OH- ions is thus: [OH-] = (moles of OH- ions at start) – (moles of OH- ions that have reacted) [OH-] = (0.100 M x 0.050 L) – (0.0105 mol /0.052 L) [OH-] = 0.0796 M To find the pH of the solution, we can use the formula: pH = 14 – pOH pH = 14 – (-log[OH-]) pH = 14 – (-log[0.0796]) pH = 14 – 1.098 pH = 12.9 Since we have added slightly more acid than would be needed to react with all of the KOH (that is, we have added a bit more than 0.100*50=5 millimoles of acid), the pH will be slightly lower than 14, but close, and the best choice in the list is C (1.27).

A 5000-ML Sample of 0