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Chapter 4: Problem 149

A 0.500 -L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0 -mL aliquot and adding 50.0 \(\mathrm{mL}\) of 0.213 \(\mathrm{M}\) NaOH. After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required 13.21 \(\mathrm{mL}\) of 0.103 \(\mathrm{M}\) \(\mathrm{HCl}\) for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) . Sulfuric acid has two acidic hydrogens.

Short Answer

Expert verified
The molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is 0.00929 M.

Step by step solution

01

Calculate the moles of excess OH- ions

First, we find out the moles of excess \(\mathrm{OH}^{-}\) ions after the reaction. We have the volume and molarity of \(\mathrm{HCl}\) for neutralization of \(\mathrm{OH}^{-}\). n(HCl) = V(HCl) × M(HCl) n(HCl) = Volume of HCl × Molarity of HCl n(OH-) = n(HCl) (Since HCl and OH- ions react in 1:1 ratio) n(OH-) = 13.21 mL × 0.103 M Convert mL to L: n(OH-) = 0.01321 L × 0.103 M n(OH-) = 0.00136033 moles
02

Calculate the moles of NaOH initially added

Next, we need to find out the moles of \(\mathrm{NaOH}\) initially added. n(NaOH) = V(NaOH) × M(NaOH) n(NaOH) = Volume of NaOH × Molarity of NaOH n(NaOH) = 50 mL × 0.213 M Convert mL to L: n(NaOH) = 0.050 L × 0.213 M n(NaOH) = 0.01065 moles
03

Calculate the moles of OH- ions that reacted with H2SO4

Now, we will subtract the excess \(\mathrm{OH}^{-}\) ions to find out the moles of \(\mathrm{OH}^{-}\) ions that reacted with the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Since sulfuric acid has two acidic hydrogens, it reacts with two moles of OH- ions per mole. n(OH-)_{reacted} = n(NaOH) - n(OH-)_{excess} n(OH-)_{reacted} = 0.01065 moles - 0.00136033 moles n(OH-)_{reacted} = 0.00928967 moles
04

Calculate the moles of H2SO4

We have the moles of \(\mathrm{OH}^{-}\) ions that reacted with the \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Since each mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with two moles of \(\mathrm{OH}^{-} \) ions: n(H2SO4) = n(OH-)_{reacted} / 2 n(H2SO4) = 0.00928967 moles / 2 n(H2SO4) = 0.004644835 moles Finally, now we have the moles of the original \(\mathrm{H}_{2} \mathrm{SO}_{4}\) sample.
05

Calculate the molarity of H2SO4

We have the moles and volume of the original \(\mathrm{H}_{2} \mathrm{SO}_{4}\) sample. M(H2SO4) = n(H2SO4) / V(H2SO4) M(H2SO4) = 0.004644835 moles / 0.500 L M(H2SO4) = 0.00928967 M The molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is 0.00929 M.

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Most popular questions from this chapter

You made 100.0 mL of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 80.0 mL left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take 2.00 mL of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of 3.407 g. What was the concentration of the original lead(II) nitrate solution?

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