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A student dissolves 0.0100 mole of an unknown weak base in \(100.0 \mathrm{mL}\) water and titrates the solution with \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) After \(40.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) was added, the \(\mathrm{pH}\) of the resulting solution was \(8.00 .\) Calculate the \(K_{\mathrm{b}}\) value for the weak base.

Short Answer

Expert verified
The \(K_b\) value for the weak base is \(6.67 \times 10^{-8}\).

Step by step solution

01

Calculate the moles of the weak base and HNO3

To start, let's first determine the moles of the weak base and the moles of HNO3 added. We are given that there are 0.0100 moles of weak base, and that the concentration of HNO3 is 0.100 M. For HNO3, we can calculate the moles using the volume of the solution and its concentration: Moles of HNO3 = Volume × Concentration Moles of HNO3 = 40.0 mL × 0.100 M = 0.0040 moles (since 1L = 1000mL, 40mL = 0.040L)
02

Calculate the moles of the weak base and its conjugate acid after titration

During the titration, the weak base will react with HNO3, and the resulting solution will contain the conjugate acid of the weak base. Since 0.0100 moles of the base react with 0.0040 moles of HNO3, we can calculate the moles of each species after the reaction: Moles of weak base after titration = Initial moles of weak base - moles of HNO3 reacted = 0.0100 - 0.0040 = 0.0060 moles Moles of conjugate acid = Moles of HNO3 reacted = 0.0040 moles
03

Calculate the concentration of OH- ions using the pH

After titration, the pH of the resulting solution is given as 8.00. We can use this to calculate the pOH and then find the concentration of OH- ions: pOH = 14 - pH pOH = 14 - 8.00 = 6.00 Now we can calculate the concentration of OH- ions using the pOH: [OH-] = 10^(-pOH) = 10^(-6.00) = 1.00 x 10^(-6) M
04

Calculate the concentrations of the weak base and conjugate acid

Now we need to determine the concentrations of the weak base and its conjugate acid. We have a total volume of 140.0 mL after mixing (100.0 mL weak base solution + 40.0 mL HNO3 solution). Concentration of weak base = moles of weak base / total volume = 0.0060 moles / 0.140 L = 0.04286 M Concentration of conjugate acid = moles of conjugate acid / total volume = 0.0040 moles / 0.140 L = 0.02857 M
05

Calculate the Kb value for the weak base

Now, we can use the concentrations of weak base, conjugate acid, and OH- ions to determine the Kb value for the weak base. The reaction can be expressed as: Weak base + H2O ⇌ Conjugate Acid + OH- The equilibrium expression for this reaction is: Kb = ([Conjugate Acid][OH-]) / [Weak base] Kb = (0.02857 M × 1.00 × 10^(-6) M) / (0.04286 M) Kb = 6.67 × 10^(-8) The Kb value for the weak base is 6.67 × 10^(-8).

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Key Concepts

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

Kb value calculation
In the realm of chemistry, particularly when dealing with acid-base reactions, calculating the equilibrium constant for bases, known as the 'base ionization constant' or 'Kb', is crucial for understanding the strength of a base and its behavior in solution. To find the Kb value, you must first understand that it denotes the equilibrium concentration of products over reactants for the ionization of a weak base in water.

As outlined in the exercise, once the moles of the base and titrant acid (HNO3) are known, and after considering the titration's stoichiometry, we can calculate the remaining concentration of the weak base and its conjugate acid. Using the pH provided, we determine the pOH, from which we calculate the hydroxide ion concentration [OH-]. With these concentrations, the Kb is simply the product of the concentrations of the conjugate acid and the OH- ions divided by the concentration of the weak base. It’s worth noting that larger Kb values correspond to stronger bases, since they ionize more extensively to form OH- ions in solution.
acid-base titration
Acid-base titration is a fundamental technique used in chemistry to determine the concentration of an unknown acid or base. In this analytical procedure, a known volume and concentration of an acid is carefully added to a base, or vice versa, until the reaction reaches its end point, which is often indicated by a color change due to an indicator or by measuring the pH.

During a weak base titration with a strong acid, like in the exercise, the weak base (B) reacts with the strong acid (HNO3) to form its conjugate acid (BH+). The titration aims to reach the equivalence point, where moles of acid are stoichiometrically equivalent to the moles of base. However, the exercise problem was concerned with the pH at a specific point of the titration, before reaching this equivalence point. This information, combined with careful calculations and knowledge of the reaction's stoichiometry, allows chemists to deduce the concentration of all species in solution and, ultimately, the base's Kb value.
pH and pOH calculations
Understanding the pH and pOH scales is essential in chemistry for describing the acidity or basicity of a solution. pH stands for 'power of hydrogen' and measures the hydrogen ion (H+) concentration in a solution, while pOH does the same for hydroxide ions (OH-). The pH scale typically runs from 0 (very acidic) to 14 (very basic), with 7 being neutral.

To relate pH and pOH, we use the equation: pH + pOH = 14. In the context of weak base titration, determining the pOH from the given pH allows us to calculate the concentration of OH- using the formula [OH-] = 10^(-pOH). Understanding these calculations is not only important for finding the concentration of ions in solution but also for progressing to further necessary calculations like determining the Kb of a base, as they provide a starting point for the mathematical journey towards comprehensive acid-base equilibrium analysis.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) after 0.010 mole of gaseous \(\mathrm{HCl}\) is added to \(250.0 \mathrm{mL}\) of each of the following buffered solutions. a. \(0.050 M \mathrm{NH}_{3} / 0.15 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) b. \(0.50 M \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?

A friend asks the following: "Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like NaOH is added, the HA reacts with the OH - to form A Thus the amount of acid (HA) is decreased, and the amount of base \(\left(\mathrm{A}^{-}\right)\) is increased. Analogously, adding HCl to the buffered solution forms more of the acid (HA) by reacting with the base \(\left(\mathrm{A}^{-}\right)\). Thus how can we claim that a buffered solution resists changes in the pH of the solution?" How would you explain buffering to this friend?

Derive an equation analogous to the Henderson-Hasselbalch equation but relating \(\mathrm{pOH}\) and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+}\).

a. Calculate the pH of a buffered solution that is 0.100 \(M\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 M\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\) b. Calculate the \(\mathrm{pH}\) after \(20.0 \%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q)$$ to calculate the pH. c. Do the same as in part b, but use the following equilibrium to calculate the pH: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q)\) d. Do your answers in parts \(b\) and c agree? Explain.

Repeat the procedure in Exercise \(61,\) but for the titration of \(25.0 \mathrm{mL}\) of 0.100 \(\mathrm{M}\) \(\mathrm{HNO}_{3}\) with 0.100 \(\mathrm{M}\) \(\mathrm{NaOH}\).

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