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Chapter 17: Problem 76

Sulfuric acid is a diprotic acid, strong in the first ionization step and weak in the second \(\left(K_{\mathrm{a}_{2}}=1.1 \times 10^{-2}\right)\) By using appropriate calculations, determine whether it is feasible to titrate \(10.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) to two distinct equivalence points with \(0.100 \mathrm{M} \mathrm{NaOH}\)

Short Answer

Expert verified
Yes, it is feasible to titrate \(H_{2}SO_{4}\) to two distinct equivalence points with \(0.100 \mathrm{M} \mathrm{NaOH}\), provided there is at least 0.002 L (\(2.00 \mathrm{mL}\)) of the \(\mathrm{NaOH}\) solution available.

Step by step solution

01

calculation for the first equivalence point

To calculate the amount of sodium hydroxide \( \mathrm{NaOH} \) required to reach the first equivalence point, we'll compute the number of moles of sulfuric acid \( \mathrm{H}_{2} \mathrm{SO}_{4} \) present, and assume that it will react completely with the same number of moles of \( \mathrm{NaOH} \). Remember that the solution volume should be in liters when calculating molarity. Hence, \n \( \text{Moles of } \mathrm{H}_{2} \mathrm{SO}_{4} = 0.100 \times 10.00/1000 = 0.001 \text{ moles} \). This amount of sulfuric acid will need the exact number of moles of \( \mathrm{NaOH} \) to fully react, hence we also need 0.001 moles of \( \mathrm{NaOH} \). This is the first equivalence point.
02

calculation for the second equivalence point

To calculate the amount of \( \mathrm{NaOH} \) required to reach the second equivalence point, we consider the number of moles of \( \mathrm{HSO}_{4}^{-} \) ion formed after the first stage of ionization, which will be equal to the initial number of moles of \( \mathrm{H}_{2} \mathrm{SO}_{4} \). Hence, we will need an additional 0.001 moles of \( \mathrm{NaOH} \) to fully react with the \( \mathrm{HSO}_{4}^{-} \) ions. This means a total of \( 0.001 + 0.001 = 0.002 \) moles of \( \mathrm{NaOH} \) is required for full titration.
03

Conclusion

We have a 0.100 M solution of \( \mathrm{NaOH} \) which we can use to titrate the \( \mathrm{H}_{2} \mathrm{SO}_{4} \). To confirm that we have enough \( \mathrm{NaOH} \), we should calculate how many moles of \( \mathrm{NaOH} \) we have. We can rearrange the molarity equation, \( M = mol/L \), to find \( mol = M \times L \). So, \( mol_{\mathrm{NaOH}} = 0.100 \times volume \). Since we don’t have any information about volume of \( \mathrm{NaOH} \), we can’t finally confirm whether there is enough \( \mathrm{NaOH} \), but if there is at least 0.002 L (\( 2.00 \mathrm{ mL} \)) of \( \mathrm{NaOH} \), it would be possible to titrate the \( \mathrm{H}_{2} \mathrm{SO}_{4} \) to two distinct equivalence points.

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Most popular questions from this chapter

In 1922 Donald D. van Slyke ( J. Biol. Chem., 52, 525) defined a quantity known as the buffer index: \(\beta=\mathrm{d} C_{\mathrm{b}} / \mathrm{d}(\mathrm{pH}),\) where \(\mathrm{d} C_{\mathrm{b}}\) represents the increment of moles of strong base to one liter of the buffer. For the addition of a strong acid, he wrote \(\beta=-\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH})\) By applying this idea to a monoprotic acid and its conjugate base, we can derive the following expression: \(\beta=2.303\left(\frac{K_{w}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\frac{\mathrm{CK}_{\mathrm{a}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left(\mathrm{K}_{\mathrm{a}}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\right)^{2}}\right)\) where \(C\) is the total concentration of monoprotic acid and conjugate base. (a) Use the above expression to calculate the buffer index for the acetic acid buffer with a total acetic acid and acetate ion concentration of \(2.0 \times 10^{-2}\) and a \(\mathrm{pH}=5.0\) (b) Use the buffer index from part (a) and calculate the \(\mathrm{pH}\) of the buffer after the addition of of a strong acid. (Hint: Let \(\left.\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH}) \approx \Delta C_{\mathrm{a}} / \Delta \mathrm{pH} .\right)\) (c) Make a plot of \(\beta\) versus \(\mathrm{pH}\) for a \(0.1 \mathrm{M}\) acetic acid buffer system. Locate the maximum buffer index as well as the minimum buffer indices.

Calculate the \(\mathrm{pH}\) at the points in the titration of \(25.00 \mathrm{mL}\) of 0.160 M HCl when (a) 10.00 mL and (b) 15.00 mL of 0.242 M KOH have been added.

Explain why the volume of \(0.100 \mathrm{M} \mathrm{NaOH}\) required to reach the equivalence point in the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HA is the same regardless of whether HA is a strong or a weak acid, yet the \(\mathrm{pH}\) at the equivalence point is not the same.

For the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(0.100 \mathrm{M} \mathrm{HCl},\) calculate the \(\mathrm{pOH}\) at a few representative points in the titration, sketch the titration curve of pOH versus volume of titrant, and show that it has exactly the same form as Figure \(17-9 .\) Then, using this curve and the simplest method possible, sketch the titration curve of pH versus volume of titrant.

What concentration of ammonia, \(\left[\mathrm{NH}_{3}\right],\) should be present in a solution with \(\left[\mathrm{NH}_{4}^{+}\right]=0.732 \mathrm{M}\) to produce a buffer solution with \(\mathrm{pH}=9.12 ?\) For \(\mathrm{NH}_{3}\) \(K_{\mathrm{h}}=1.8 \times 10^{-5}\)
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