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If \(\mathrm{pK}_{b}\) for \(\mathrm{CN}\) - at \(25^{\circ} \mathrm{C}\) is \(4.7\), the \(\mathrm{pH}\) of \(0.5 \mathrm{M}\) aqueous \(\mathrm{NaCN}\) solution is (a) 10 (b) \(11.5\) (c) 11 (d) 12

Short Answer

Expert verified
The pH of the solution is 11.5, which corresponds to option (b).

Step by step solution

01

Understand the relationship between pKb and pKa

To find the pH of the solution, we need to relate the base dissociation constant (pKb) to the acid dissociation constant (pKa). Since CN鈦 acts as a base and HCN is its conjugate acid, we use: \( \mathrm{pK}_{w} = \mathrm{pK}_{a} + \mathrm{pK}_{b} \). At 25掳C, \( \mathrm{pK}_{w} = 14 \). Given \( \mathrm{pK}_{b} = 4.7 \), calculate \( \mathrm{pK}_{a} \) as follows: \( \mathrm{pK}_{a} = 14 - 4.7 = 9.3 \).
02

Find the concentration of OH鈦 ions

The molarity of NaCN is 0.5 M. In solution, CN鈦 forms OH鈦 ions: \( \mathrm{CN}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^{-} \). Use the relationship \( \mathrm{pH} + \mathrm{pOH} = 14 \) and the fact that \( \mathrm{pOH} = \dfrac{1}{2} (\mathrm{pK}_{b} - \log [\text{base}]) \) to find \( \mathrm{pOH} \). Substitute the Base concentration = 0.5 M: \( \mathrm{pOH} = \dfrac{1}{2} (4.7 - \log 0.5) \).
03

Evaluate the expression for pOH

Calculate \( \log 0.5 \), knowing that \( \log 0.5 = -0.3010 \), then \( \mathrm{pOH} = \dfrac{1}{2} (4.7 + 0.3010) = \dfrac{1}{2} \times 5.001 = 2.5005 \).
04

Calculate the pH from the pOH

Since \( \mathrm{pH} + \mathrm{pOH} = 14 \), we get \( \mathrm{pH} = 14 - 2.5005 = 11.5 \).
05

Choose the correct answer

From the calculated \( \mathrm{pH} \), the value matches option (b), which is 11.5.

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Key Concepts

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

pKa and pKb relationship
Understanding the relationship between pKa and pKb is important while calculating pH in solutions involving weak acids and bases. These terms are measures of the strengths of acids and bases:
  • pKa denotes the acid dissociation constant for a conjugate acid.
  • pKb denotes the base dissociation constant for a conjugate base.
At a standard temperature of 25掳C, the ion product constant for water (\[\mathrm{pK}_w = 14\]) helps relate pKa and pKb.
Simply, \[\mathrm{pK}_w = \mathrm{pK}_a + \mathrm{pK}_b\].
This means any given conjugate acid-base pair has intertwined pKa and pKb values, facilitating the easy shift between the two constants when provided with one and the need to calculate the other. For example, if you know the pKb of \(\mathrm{CN}^{-}\) is 4.7 as in this example, then its conjugate, HCN, has a pKa calculated by \(\mathrm{pK}_a = 14 - \mathrm{pK}_b = 9.3\). Understanding this balance, more so from the perspective of the stable equilibrium, helps students easily compute necessary ionic values in solution chemistry.
dissociation constants
Dissociation constants are numbers that provide practical insights into the equilibrium between a molecule and its dissociated ions in solution. They help in deciphering details about the extent a compound separates into ions:
  • Ka: Refers to acid dissociation constant, capturing the equilibrium strength for an acid converting into its constituent ions.
  • Kb: Stands for base dissociation constant, which values the equilibrium of a base accepting protons to transform into its conjugate acid.
The inverse log of these constants gives us signs like pKa and pKb, as used in everyday chemistry practices. They aid in determining the direction and extents like \[\mathrm{CN}^{-}\] ionizing to the bent HCN equilibrium as observed in the exercise.
Consultation with these constants quickly informs on the balance state an acid or base often adopts, helping project pH derivations from known concentrations and progressive ionization tendencies as demonstrated.
base dissociation in water
Base dissociation in water is an essential concept when dealing with alkaline solutions and often comes to play when predicting the pH of solutions, such as sodium cyanide (NaCN). The reaction involves a base, such as \[\mathrm{CN}^{-}\], interacting with water and establishing an equilibrium:
\[\mathrm{CN}^{-} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^{-}\].
Here, water donates a proton to the base, producing hydroxide ions (\[\mathrm{OH}^{-}\]), which elevate the alkalinity, thus affecting the pH.
This dissociation is captured using Kb. Using formulas like \(pH + pOH = 14\), and \(pOH = \frac{1}{2}(pK_b - \log [\text{base}])\), you calculate the hydroxide concentration, allowing easy passage to pH calculations.
  • The presence of \[\mathrm{OH}^{-}\] impacts the solution's pH by increasing it, making it more basic.
In scenarios like NaCN's dissociation, understanding how these waters simplest components transform, provide clarity on hydrogen and hydroxide concentration balances in an aqueous environment.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\), the solubility product of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(1.0 \times 10^{-11}\). At which \(\mathrm{pH}\), will \(\mathrm{Mg}^{2+}\) ions start precipitating in the form of \(\mathrm{Mg}(\mathrm{OH})_{2}\) from a solution of \(0.001\) M \(\mathrm{Mg}^{2+}\) ions? (a) 9 (b) 10 (c) 11 (d) 8

Hydrolysis constant \(\mathrm{K}_{\mathrm{A}}\) and \(\mathrm{K}_{\mathrm{B}}\) of two salts of weak acids HA and \(\mathrm{HB}\) are \(10^{-8}\) and \(10^{-6}\) respectively. If the dissociation constant of third acid \(\mathrm{HC}\) is \(10^{-2}\). The order of acidic strengths of three acids will be (a) \(\mathrm{HA}>\mathrm{HB}>\mathrm{HC}\) (b) \(\mathrm{HB}>\mathrm{HA}>\mathrm{HC}\) (c) \(\mathrm{HC}>\mathrm{HA}>\mathrm{HB}\) (d) \(\mathrm{HA}=\mathrm{HB}=\mathrm{HC}\)

The solubility product of a salt having general formula \(\mathrm{MX}_{2}\) in water is, \(4 \times 10^{-12}\) \([2005]\) The concentration of \(\mathrm{M}^{2+}\) ions in the aqueous solution of the salt is (a) \(1.6 \times 10^{-4} \mathrm{M}\) (b) \(2.0 \times 10^{-6} \mathrm{M}\) (c) \(1.0 \times 10^{-4} \mathrm{M}\) (c) \(4.0 \times 10^{-10} \mathrm{M}\)

An acid-base indicator has \(\mathrm{K}_{\mathrm{a}}=3.0 \times 10^{-5} .\) The acid form of the indicator is red and the basic form is blue. The \(\left[\mathrm{H}^{+}\right]\)required to change the indicator from \(75 \%\) red to \(75 \%\) blue is (a) \(8 \times 10^{-5} \mathrm{M}\) (b) \(9 \times 10^{-5} \mathrm{M}\) (c) \(1 \times 10^{-5} \mathrm{M}\) (d) \(3 \times 10^{-4} \mathrm{M}\)

$$ \begin{aligned} &\text { Match the following }\\\ &\begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (a) } \mathrm{FeCl}_{3} \text { solution (aqueous) } & \text { (p) } \mathrm{pH}<7 \\ \text { (b) } \mathrm{CH}_{3} \text { COONa solution (aqueous) } & \text { (q) } \mathrm{pH}>7 \\ \text { (c) Mixture of } 0.1 \mathrm{M} \text { acetic acid and } & \text { (r) } \mathrm{pH}=7 \\ 0.1 \mathrm{M} \text { sodium acetate (aqueous) } & \\ \begin{array}{ll} \text { (d) } 0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONH}_{4} \text { (aqueous) } & \text { (s) acidic } \\ & \text { (t) basic } \\ \hline \end{array} \end{array} \end{aligned} $$

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