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Chapter 17: Problem 41

How many milliliters of \(0.0750 \mathrm{M} \mathrm{KOH}\) are required to titrate each of the following solutions to the equivalence point: (a) \(30.0 \mathrm{~mL}\) of \(0.0900 \mathrm{M} \mathrm{HCOOH}\), (b) \(45.0 \mathrm{~mL}\) of \(0.0750 \mathrm{M} \mathrm{HNO}_{3},(\mathbf{c}) 50.0 \mathrm{~mL}\) of a solution that contains \(3.00 \mathrm{~g}\) of HBr per liter?

Short Answer

Expert verified
(a) 36.0 mL KOH, (b) 45.0 mL KOH, (c) 24.67 mL KOH.

Step by step solution

01

Determine moles of HCOOH

First, calculate the moles of \mathrm{HCOOH} in the solution. Use the formula \[ \text{moles} = \text{concentration (M)} \times \text{volume (L)} \]For \(a\), the concentration is \(0.0900 \, \mathrm{M}\) and the volume is \(30.0 \, \mathrm{mL} = 0.0300 \, \mathrm{L}\).\[ \text{moles of } \mathrm{HCOOH} = 0.0900 \, \mathrm{M} \times 0.0300 \, \mathrm{L} = 0.00270 \, \text{moles} \]
02

Calculate KOH required for HCOOH

The reaction between \mathrm{KOH} and \mathrm{HCOOH} is a 1:1 molar reaction, meaning one mole of \mathrm{KOH} reacts with one mole of \mathrm{HCOOH}.Thus, \textbf{the moles of \mathrm{KOH}} required is also \(0.00270\, \text{moles}\).Using the concentration of \(0.0750 \, \mathrm{M} \, \mathrm{KOH}\), we calculate the volume required:\[ \text{Volume (L)} = \frac{\text{moles}}{\text{concentration}} = \frac{0.00270 \, \text{moles}}{0.0750 \, \mathrm{M}} = 0.0360 \, \mathrm{L} \]Convert to milliliters:\[ 0.0360 \, \mathrm{L} \times 1000 = 36.0 \, \mathrm{mL} \]
03

Determine moles of HNO3

Calculate the moles of \mathrm{HNO_3} in the solution. For \(b\), the concentration is \(0.0750 \, \mathrm{M}\) and the volume is \(45.0 \, \mathrm{mL} = 0.0450 \, \mathrm{L}\).\[ \text{moles of } \mathrm{HNO_3} = 0.0750 \, \mathrm{M} \times 0.0450 \, \mathrm{L} = 0.00338 \, \text{moles} \]
04

Calculate KOH required for HNO3

Since \mathrm{KOH} and \mathrm{HNO_3} also react in a 1:1 molar fashion, the moles of \mathrm{KOH} required is \(0.003375 \, \,\text{moles}\).The volume of \mathrm{KOH} needed is:\[ \text{Volume (L)} = \frac{0.00338 \, \text{moles}}{0.0750 \, \mathrm{M}} = 0.0450 \, \mathrm{L} \]Convert to milliliters:\[ 0.0450 \, \mathrm{L} \times 1000 = 45.0 \, \mathrm{mL} \]
05

Determine moles of HBr

For \(c\), calculate the concentration of \mathrm{HBr} from the given mass per liter. Molar mass of \mathrm{HBr} = 1 + 80 = 81 \, \text{g/mol}.Concentration of \mathrm{HBr}:\[ = \frac{3.00 \, \text{g}}{81 \, \text{g/mol}}= 0.0370 \, \text{M} \]Calculate the moles of \mathrm{HBr} using this concentration:\[ \text{moles of } \mathrm{HBr} = 0.0370 \, \text{M} \times 0.0500 \, \mathrm{L} = 0.00185 \, \text{moles} \]
06

Calculate KOH required for HBr

The moles of \mathrm{KOH} needed for \mathrm{HBr} is equivalent to its moles: \(0.00185 \, \text{moles}\).Volume of \mathrm{KOH}:\[ \text{Volume (L)} = \frac{0.00185 \, \text{moles}}{0.0750 \, \mathrm{M}} = 0.02467 \, \mathrm{L} \]Convert to milliliters:\[ 0.02467 \, \mathrm{L} \times 1000 = 24.67 \, \mathrm{mL} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid-Base Reactions
In a titration involving acid-base reactions, the goal is to find the exact amount of an acid that can neutralize a base, or vice versa. This process helps us determine the concentration of one solution using a solution of known concentration. The key reaction in the given exercise involves potassium hydroxide (KOH) and different acids like HCOOH (formic acid), HNO3 (nitric acid), and HBr (hydrobromic acid).

A titration reaches its endpoint at the equivalence point, where the moles of acid equal the moles of base. The chemical reaction involves the hydrogen ions from the acid reacting with the hydroxide ions from the base to form water and a salt. For example:
  • KOH + HCOOH 鈫� H2O + KCOOH
  • KOH + HNO3 鈫� H2O + KNO3
  • KOH + HBr 鈫� H2O + KBr
Understanding the stoichiometry of the reactions is essential as they guide the calculations required to find the amounts needed during titration.
Molarity Calculations
Molarity is a measure of the concentration of a solution, given in moles of solute per liter of solution (mol/L). Accurate molarity calculations are crucial in titrations as the solution鈥檚 concentration directly affects the volume measurement when reacting with another substance.

In this exercise, you calculate the amount of potassium hydroxide (KOH) needed by first determining how many moles of acid are present in the solution. Use the formula:
  • \[\text{moles of solute} = \text{molarity} \times \text{volume in liters}\]
Then, calculate the volume of KOH needed to reach the equivalence point using:
  • \[\text{Volume of solution} = \frac{\text{moles required}}{\text{molarity of } KOH}\]
For example, if you have 0.0900 M of HCOOH in 30.0 mL, convert the volume to liters, calculate the moles, and then find how much 0.0750 M KOH is needed to neutralize the acid.
Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. The stoichiometric coefficients of a balanced chemical equation tell us the proportions in which chemicals react. In acid-base titrations, it's typically a 1:1 ratio unless specified differently by the reaction equation.

For titrating HCOOH, HNO3, and HBr with KOH, the reactions happen at a 1:1 ratio. This means that one mole of KOH neutralizes one mole of acid. Here's a step-by-step on using stoichiometry in titration:
  • Calculate the moles of the acid present using molarity and volume.
  • Assuming the reaction is a 1:1 ratio, the moles of KOH needed will be equal to the acid's moles.
  • Finally, determine the moles of KOH in the required volume of solution through its molarity.
Once stoichiometry is understood, you can confidently calculate how much titrant is needed for any given reaction, assuring the balance and completion of a chemical reaction.

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Most popular questions from this chapter

Lead(II) carbonate, \(\mathrm{PbCO}_{3}\), is one of the components of the passivating layer that forms inside lead pipes. (a) If the \(K_{i p}\) for \(\mathrm{PbCO}_{3}\) is \(7.4 \times 10^{-14}\) what is the molarity of \(\mathrm{Pb}^{2+}\) in a saturated solution of lead(II) carbonate? (b) What is the concentration in ppb of \(\mathrm{Pb}^{2+}\) ions in a saturated solution? (c) Will the solubility of \(\mathrm{PbCO}_{3}\) increase or decrease as the \(\mathrm{pH}\) is lowered? (d) The EPA threshold for acceptable levels of lead ions in water is 15 ppb. Does a saturated solution of lead(II) carbonate produce a solution that exceeds the EPA limit?

Furoic acid \(\left(\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4} \mathrm{at}\) \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of \((\mathbf{a})\) a solution formed by adding \(30.0 \mathrm{~g}\) of furoic acid and \(25.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.300 \mathrm{~L}\) of solution, \((\mathbf{b})\) a solution formed by mixing \(20.0 \mathrm{~mL}\). of \(0.200 \mathrm{M}\) \(\mathrm{HC}_{\mathrm{s}} \mathrm{H}_{3} \mathrm{O}_{3}\) and \(30.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and diluting the total volume to \(125 \mathrm{~mL},(\mathbf{c})\) a solution prepared by adding \(25.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) solution to \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{MHC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\)

Consider a beaker containing a saturated solution of CaF \(_{2}\) in equilibrium with undissolved \(\mathrm{CaF}_{2}(s)\). Solid \(\mathrm{CaCl}_{2}\) is then added to the solution. (a) Will the amount of solid \(\mathrm{CaF}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Ca}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of F ions in solution increase or decrease?

A student who is in a great hurry to finish his laboratory work decides that his qualitative analysis unknown contains a metal ion from group 4 of Figure \(17.23 .\) He therefore tests his sample directly with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4},\) skipping earlier tests for the metal ions in groups \(1,2,\) and 3 . He observes a precipitate and concludes that a metal ion from group 4 is indeed present. Why is this possibly an erroneous conclusion?

Aspirin has the structural formula CC(=O)Oc1ccccc1C(=O)O At body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{a}\) for aspirin equals \(3 \times 10^{-5}\). If two aspirin tablets, each having a mass of \(325 \mathrm{mg}\), are dissolved in a full stomach whose volume is \(1 \mathrm{~L}\) and whose \(\mathrm{pH}\) is \(2,\) what percent of the aspirin is in the form of neutral molecules?
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