Question

The heat of neutralization, \(\Delta H_{\text {neut }},\) can be defined as the amount of heat released (or absorbed), \(q,\) per mole of acid (or base) neutralized. \(\Delta H_{\text {neut }}\) for nitric acid is \(-52 \mathrm{~kJ} / \mathrm{mol}\) \(\mathrm{HNO}_{3} .\) At \(27.3^{\circ} \mathrm{C}, 50.00 \mathrm{~mL}\) of \(0.743 \mathrm{M} \mathrm{HNO}_{3}\) is neutral- ized by \(1.00 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\) in a coffee-cup calorimeter. (a) How many \(\mathrm{mL}\) of \(\mathrm{Sr}(\mathrm{OH})_{2}\) were used in the neutralization? (b) What is the final temperature of the resulting solution? (Use the assumptions in Ouestion \(11 .)\)

Short answer

The volume of strontium hydroxide used in the reaction is 18.6 mL, and the final temperature of the resulting solution is 31.3°C.

Step-by-step solution

(Step 1: Calculate moles of nitric acid)

We are given the initial volume and concentration of nitric acid. We will use the equation: Moles of nitric acid = Volume x Concentration Moles of \(\mathrm{HNO}_3\) = \((50.0\,\mathrm{mL})(0.743\,\mathrm{M})\) Moles of \(\mathrm{HNO}_3\) = \(0.0372\,\mathrm{mol}\)

(Step 2: Determine moles of strontium hydroxide needed)

The balanced chemical equation for the neutralization of nitric acid with strontium hydroxide can be written as: \(\mathrm{2HNO}_3 + \mathrm{Sr(OH)}_2 \longrightarrow \mathrm{Sr(NO}_3)_2+\mathrm{2H_2O}\) From the balanced equation, we can see that 2 moles of nitric acid react with 1 mole of strontium hydroxide. So, we can divide moles of \(\mathrm{HNO}_3\) by 2 to find moles of \(\mathrm{Sr}(\mathrm{OH})_2\): Moles of \(\mathrm{Sr}(\mathrm{OH})_2\) = \(0.0372\,\mathrm{mol}/2 = 0.0186\,\mathrm{mol}\)

(Step 3: Calculate the volume of strontium hydroxide used)

We are given the concentration of strontium hydroxide and we have determined the moles needed for neutralization. We can use the equation: Volume = Moles / Concentration Volume of \(\mathrm{Sr}(\mathrm{OH})_2\) = \(0.0186\,\mathrm{mol}/1.00\,\mathrm{M} = 0.0186\,\mathrm{L}\) Convert the volume to milliliters: Volume of \(\mathrm{Sr}(\mathrm{OH})_2\) = \(0.0186\,\mathrm{L}\times1000\,\mathrm{mL/L} = 18.6\,\mathrm{mL}\)

(Step 4: Calculate the heat released during the reaction)

We are given the heat of neutralization for nitric acid as \(-52\,\mathrm{kJ/mol}\) and we have determined the moles of nitric acid reacted. Heat released \(q\) = Moles of \(\mathrm{HNO}_3\) \(\times\) \(\Delta H_{\text {neut }}\) \(q\) = \((0.0372\,\mathrm{mol})(-52\,\mathrm{kJ/mol})\) \(q\) = \(-1.93\,\mathrm{kJ}\)

(Step 5: Calculate the final temperature of the resulting solution)

The heat capacity of the resulting solution can be approximated as the heat capacity of water, which is \(4.184\,\mathrm{J/g\,^\circ C}\). We are asked to use assumptions in Question 11, which are: (a) The density of the solution is approximately 1.00 g/mL. (b) The total volume of the solution is approximately 100.0 mL. (c) The specific heat capacity of the solution is approximately the same as that of pure water. First, we convert the released heat to Joules: \(q\) = \(((1.93\,\mathrm{kJ})(1000\,\mathrm{J/kJ)})=-1930\,\mathrm{J}\) Now, find the mass of the resulting solution: Mass = Volume x Density Mass = \((100.0\,\mathrm{mL})(1.00\,\mathrm{g/mL})=100.0\,\mathrm{g}\) Now, we can use the equation: \(q = mc(T_f-T_i)\) \(-1930\,\mathrm{J} = (100.0\,\mathrm{g})(4.184\,\mathrm{J/g\,^\circ C})(T_f-27.3^{\circ}\mathrm{C})\) Solve for the final temperature, \(T_f\): \(T_f = (-1930\,\mathrm{J}/(100.0\,\mathrm{g}\times 4.184\,\mathrm{J/g\,^\circ C}))+27.3^{\circ}\mathrm{C}\) \(T_f = 31.3^{\circ}\mathrm{C}\) The final temperature of the resulting solution is \(31.3^{\circ}\mathrm{C}\).