Question

When \(25.00 \mathrm{~mL}\) of \(\mathrm{HNO}_{3}\) are titrated with \(\mathrm{Sr}(\mathrm{OH})_{2}\), \(58.4 \mathrm{~mL}\) of a \(0.218 \mathrm{M}\) solution are required. (a) What is the \(\mathrm{pH}\) of \(\mathrm{HNO}_{3}\) before titration? (b) What is the \(\mathrm{pH}\) at the equivalence point? (c) Calculate \(\left[\mathrm{NO}_{3}^{-}\right]\) and \(\left[\mathrm{Sr}^{2+}\right]\) at the equivalence point. (Assume that volumes are additive.)

Short answer

Answer: The pH before titration is approximately 0.60, the pH at the equivalence point is 7, and the concentrations of NO3- and Sr2+ ions at the equivalence point are approximately 0.305 M and 0.152 M, respectively.

Step-by-step solution

Write the balanced chemical equation for the reaction.

The reaction between HNO3 and Sr(OH)2 can be written as: $$ 2\mathrm{HNO}_{3}(aq) + \mathrm{Sr}(\mathrm{OH})_{2}(aq) \rightarrow \mathrm{Sr}(\mathrm{NO}_{3})_{2}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l) $$

Calculate the initial concentration of HNO3.

The given information is that 25.00 mL of HNO3 was titrated with 58.4 mL of a 0.218 M solution of Sr(OH)2. Let's denote the initial concentration of HNO3 as [HNO3]\(_{initial}\). From the stoichiometry of the balanced chemical equation, we can see that 2 moles of HNO3 react with 1 mole of Sr(OH)2. Using this information, we can write: $$ \frac{2}{1} = \frac{\left[\mathrm{HNO}_{3}\right]_{initial} \cdot 25.00 mL}{0.218 M \cdot 58.4 mL} $$ Now, we can solve for \(\left[\mathrm{HNO}_{3}\right]_{initial}\): $$\left[\mathrm{HNO}_{3}\right]_{initial} = \frac{0.218 M \cdot 58.4 mL}{2 \cdot 25.00 mL} = 0.25336 M$$

Calculate the pH of HNO3 before titration.

HNO3 is a strong acid, so it completely ionizes in water. Therefore, its initial concentration equals its initial H+ ion concentration. So for part (a), the pH before titration is given by the formula: $$\mathrm{pH} = -\log[\mathrm{H}^{+}]$$ $$\mathrm{pH} = -\log(0.25336) \approx 0.60$$

Calculate the pH at the equivalence point.

At the equivalence point, the moles of HNO3 will be equal to the moles of OH-. Since we have 2 moles of HNO3 reacting with 1 mole of Sr(OH)2, It means that both the moles of HNO3 and OH- are equal at the equivalence point. Therefore, all of the HNO3 has been neutralized and there are no H+ ions left in the solution. The pH at the equivalence point will be 7, as it is a neutral solution. So, for part (b), the pH at the equivalence point is 7.

Calculate the concentrations of NO3- and Sr2+ ions at the equivalence point.

At the equivalence point, all of the initial HNO3 has reacted with Sr(OH)2 to form Sr(NO3)2. We can use the stoichiometry of the reaction to find the concentrations of NO3- and Sr2+ ions. From the balanced chemical equation, we know that the 2 moles of \(\mathrm{NO}_{3}^{-}\) are produced for 1 mole of \(\mathrm{Sr}^{2+}\). Calculating the moles of NO3- and Sr2+ at the equivalence point: $$moles\ of\ \mathrm{Sr}^{2+} = 0.218 M \times 58.4 mL = 0.01272 mol$$ $$moles\ of\ \mathrm{NO}_{3}^{-} = 2 \times moles\ of\ \mathrm{Sr}^{2+} = 2 \times 0.01272 mol = 0.02544 mol$$ At the equivalence point, the total volume of the solution is 25.00 mL + 58.4 mL = 83.4 mL. Calculating the concentrations of NO3- and Sr2+ ions at the equivalence point: $$\left[\mathrm{NO}_{3}^{-}\right] = \frac{0.02544\ mol}{83.4\ mL} \approx 0.305 M$$ $$\left[\mathrm{Sr}^{2+}\right] = \frac{0.01272\ mol}{83.4\ mL} \approx 0.152 M$$ So, for part (c), \(\left[\mathrm{NO}_{3}^{-}\right] \approx 0.305 M\) and \(\left[\mathrm{Sr}^{2+}\right] \approx 0.152 M\) at the equivalence point.