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Chapter 5: Problem 57

A \(25.00 \mathrm{mL}\) sample of \(0.132 \mathrm{M}\) \(\mathrm{HNO}_{3}\) is mixed with \(10.00 \mathrm{mL}\) of \(0.318 \mathrm{M} \mathrm{KOH} .\) Is the resulting solution acidic, basic, or exactly neutralized?

Short Answer

Expert verified
The resulting solution is acidic.

Step by step solution

01

Calculate the moles of acid and base

The moles of HNO3 and KOH can be calculated using the given volume and molarity. The formula used for this is: Molarity = Moles/Volume. Rearranging for Moles gives: Moles = Molarity * Volume (note that the volume must be converted to liters). Hence, the moles of HNO3 and KOH are: Moles(HNO3) = 0.132 M * 0.02500 L = 0.0033 moles Moles(KOH) = 0.318 M * 0.01000 L = 0.00318 moles
02

Determine the limiting reactant

In the reaction of HNO3 with KOH, the acid and the base react in a 1:1 ratio to produce water and a salt (in this case KNO3). Therefore, we compare the number of moles of each reactant to see which one is completely consumed (the limiting reactant). From our calculations in Step 1, we know that there are more moles of HNO3 (0.0033 moles) than there are of KOH (0.00318 moles), so KOH is the limiting reactant.
03

Determine the nature of the solution

Since the KOH, which is a base, is used up completely, there will be a surplus of HNO3 in the solution. Hence, the nature of the resulting solution will be acidic as it contains excess HNO3.

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Most popular questions from this chapter

Balance these equations for disproportionation reactions. (a) $\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}$ (basic solution) (b) $\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{HSO}_{3}^{-}$ (acidic solution)

To precipitate \(\mathrm{Zn}^{2+}\) from \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}),\) add (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{MgBr}_{2} ;\) (c) \(\mathrm{K}_{2} \mathrm{CO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

The exact neutralization of \(10.00 \mathrm{mL}\) of \(0.1012 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\text { aq })\) requires \(23.31 \mathrm{mL}\) of \(\mathrm{NaOH}\). What must be the molarity of the \(\mathrm{NaOH}(\mathrm{aq}) ?\) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1)\)

Complete each of the following as a net ionic equation, indicating whether a precipitate forms. If no reaction occurs, so state. (a) $\mathrm{Na}^{+}+\mathrm{Br}^{-}+\mathrm{Pb}^{2+}+2 \mathrm{NO}_{3}^{-} \longrightarrow$ (b) $\mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-}+\mathrm{Cu}^{2+}+\mathrm{SO}_{4}^{2-} \longrightarrow$ (c) $\mathrm{Fe}^{3+}+3 \mathrm{NO}_{3}^{-}+\mathrm{Na}^{+}+\mathrm{OH}^{-} \longrightarrow$

Which aqueous solution has the greatest \(\left[\mathrm{H}^{+}\right]:\) (a) \(0.011 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} ;\) (b) \(0.010 \mathrm{M} \mathrm{HCl} ;\) (c) \(0.010 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(1.00 \mathrm{M} \mathrm{NH}_{3} ?\) Explain your choice.
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