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Chapter 5: Problem 52

How many milliliters of \(0.0844 \mathrm{MBa}(\mathrm{OH})_{2}\) are required to titrate \(50.00 \mathrm{mL}\) of \(0.0526 \mathrm{M} \mathrm{HNO}_{3} ?\)

Short Answer

Expert verified
You would require approximately 15.6 mL of \(0.0844 \mathrm{MBa}(\mathrm{OH})_{2}\) to titrate 50.00 mL of \(0.0526 \mathrm{M} \mathrm{HNO}_{3}\).

Step by step solution

01

Analyze the chemical reaction

In the reaction between nitric acid and barium hydroxide, 1 mole of barium hydroxide reacts with 2 moles of nitric acid. You can write it down as follows: \(Ba(OH)_2 + 2HNO_3 \rightarrow Ba(NO_3)_2 + 2H_2O\)
02

Calculate moles of nitric acid

First, calculate the moles of nitric acid (\(HNO_3\)) in the metioned volume. The moles of \(HNO_3\) can be calculated using the given concentration and volume. You can use the formula: moles = concentration (in M) * Volume(in Liters). The volume needs to be converted from mL to Liters, where 1mL = 0.001L. Hence, moles of \(HNO_3\)= \(0.0526 \mathrm{M} * 50.00 \mathrm{mL}*0.001 \mathrm{L/m}L\) = \(0.00263 \mathrm{mol}\)
03

Calculate volume of \(Ba(OH)_2\)

Since 2 moles of nitric acid react with 1 mole of barium hydroxide, the number of moles of \(Ba(OH)_2\) required to completely react with \(0.00263 \mathrm{mol}\) of \(HNO3\) will be half of that, i.e., \(0.001315 \mathrm{mol}\). Now, let V be the volume in liters of \(Ba(OH)_2\). As concentration \(C = \frac{moles}{volume}\), so rearranging for volume, we get \(V = \frac{moles}{C}\). Substituting for moles and C, we get: \(V = \frac{0.001315 \mathrm{mol}}{0.0844 \mathrm{M}} = 0.0156 \mathrm{L}\). Converting this to mL (as 1L = 1000mL), we get \(V = 0.0156 * 1000 = 15.6 \mathrm{mL}\).

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Most popular questions from this chapter

Balance these equations for redox reactions occurring in acidic solution. (a) $\mathrm{P}_{4}(\mathrm{s})+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{NO}(\mathrm{g})$ (b) $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{SO}_{4}^{2-}+\mathrm{Mn}^{2+}$ (c) $\mathrm{HS}^{-}+\mathrm{HSO}_{3}^{-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ (d) $\mathrm{Fe}^{3+}+\mathrm{NH}_{3} \mathrm{OH}^{+} \longrightarrow \mathrm{Fe}^{2+}+\mathrm{N}_{2} \mathrm{O}(\mathrm{g})$

Balance these equations for reactions in acidic solution. (a) $\mathrm{IBr}+\mathrm{BrO}_{3}^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{IO}_{3}^{-}+\mathrm{Br}^{-}+\mathrm{H}_{2} \mathrm{O}$ (b) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{3}+\mathrm{Sn} \longrightarrow$ $\mathrm{NH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Sn}^{2+}$ (c) $\mathrm{As}_{2} \mathrm{S}_{3}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{S}+\mathrm{NO}$ (d) $\mathrm{H}_{5} \mathrm{IO}_{6}+\mathrm{I}_{2} \longrightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{O}$ (e) $\mathrm{S}_{2} \mathrm{F}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{S}_{8}+\mathrm{H}_{2} \mathrm{S}_{4} \mathrm{O}_{6}+\mathrm{HF}$

When aqueous sodium carbonate, \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), is treated with dilute hydrochloric acid, HCl, the products are sodium chloride, water, and carbon dioxide gas. What is the net ionic equation for this reaction?

Balance these equations for redox reactions occurring in basic solution. (a) \(\mathrm{CrO}_{4}^{2-}+\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{SO}_{3}^{2-}\) (b) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+\mathrm{N}_{2}(\mathrm{g})\) (c) \(\operatorname{Fe}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{MnO}_{4}^{-} \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{MnO}_{2}(\mathrm{s})\)

Assuming the volumes are additive, what is the \(\left[\mathrm{NO}_{3}^{-}\right]\) in a solution obtained by mixing \(275 \mathrm{mL}\) of \(0.283 \mathrm{M} \mathrm{KNO}_{3}, 328 \mathrm{mL}\) of \(0.421 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2},\) and \(784 \mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{O} ?\)
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