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Chapter 7: Problem 70

Which of the following statements(s) is (are) correct? (a) The pH of \(1.0 \times 10^{-8} \mathrm{M}\) solution of \(\mathrm{HCl}\) is 8 (b) The conjugate base of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)is \(\mathrm{HPO}_{4}^{2-}\) (c) Autoprotolysis constant of water increases with temperature (d) When a solution of a weak monoprotic acid is titrated against a strong base, at half-neutralisation pointpH \(=(1 / 2) \mathrm{p} K_{a}\).

Short Answer

Expert verified
Only statements (b) and (c) are correct.

Step by step solution

01

Analyzing statement (a)

For statement (a), calculate the pH of a \(1.0 \times 10^{-8} \mathrm{M}\) \(\mathrm{HCl}\) solution. Since the concentration is very low, we must consider the ionization of water. The ionic product of water \(\mathrm{K}_w\) is \(1.0 \times 10^{-14}\). In neutral water, \([\mathrm{H}^+] = [\mathrm{OH}^-] = 1.0 \times 10^{-7} \mathrm{M}\). Hence, the actual \([\mathrm{H}^+]\) will be \(1.0 \times 10^{-7} \mathrm{M} + 1.0 \times 10^{-8} \mathrm{M}\). Calculate the pH now: \[\mathrm{pH} = -\log_{10}(1.0 \times 10^{-7} + 1.0 \times 10^{-8})\] The pH is actually slightly less than 7, so this statement is incorrect.
02

Evaluating statement (b)

The conjugate base of an acid is formed by removing a proton \((\mathrm{H}^+)\) from the acid. \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is the given acid, so its conjugate base will be \(\mathrm{HPO}_{4}^{2-}\). Since removing a proton reduces the charge by 1, this is correct.
03

Investigating statement (c)

Consider the effect of temperature on the autoprotolysis constant \(\mathrm{K}_w\) of water. The reaction \(2 \mathrm{H}_2\mathrm{O} \leftrightarrow \mathrm{H}_3\mathrm{O}^+ + \mathrm{OH}^-\) is endothermic. Raising the temperature shifts the equilibrium to produce more \(\mathrm{H}_3\mathrm{O}^+\) and \(\mathrm{OH}^-\), therefore increasing \(\mathrm{K}_w\). Thus, this statement is correct.
04

Analyzing statement (d)

At the half-neutralization point in a titration of a weak monoprotic acid with a strong base, the concentration of the acid is equal to the concentration of its conjugate base. Hence, \([\mathrm{HA}] = [\mathrm{A}^-]\), leading to the equation \(\mathrm{pH} = \mathrm{pK}_a\), not \(\mathrm{pH} = (1/2)\mathrm{pK}_a\). This makes statement (d) incorrect.

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Key Concepts

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

pH calculation
Calculating the pH of a solution involves understanding the concentration of hydrogen ions \(\mathrm{H}^+\) in the solution. For most dilute strong acids like hydrochloric acid \(\mathrm{HCl}\), the pH calculation considers the ionization of water as well. In this exercise, the concentration of \(\mathrm{HCl}\) was given as \(1.0 \times 10^{-8} \mathrm{M}\). This small concentration is actually less than that of the hydrogen ions naturally found in pure water \(\mathrm{H}_2\mathrm{O}\).
Neutral water has a concentration of hydrogen ions (\

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Most popular questions from this chapter

What will be the resultant \(\mathrm{pH}\) when \(200 \mathrm{~mL}\) of an aqueous solution of \(\mathrm{HCl}(\mathrm{pH}=2.0)\) is mixed with \(300 \mathrm{~mL}\) of an aqueous solution of \(\mathrm{NaOH}(\mathrm{pH}=12.0) ?\)

The best indicator for detection of end point in titration of a weak acid and a strong base is : (a) methyl orange ( 3 to 4 ) (b) methyl red (5 to 6) (c) bromothymol blue ( 6 to \(7.5\) ) (d) phenolphthalein ( 8 to \(9.6\) )

The variation of equilibrium constant with temperature is given below: $$ \begin{array}{ll} {\text { Temperature }} & {\text { Equilibrium Constant }} \\ \mathrm{T}_{1}=25^{\circ} \mathrm{C} & \mathrm{K}_{1}=10 \\ \mathrm{~T}_{2}=100{ }^{\circ} \mathrm{C} & \mathrm{K}_{2}=100 \end{array} $$ The values of \(\Delta \mathrm{H}^{\circ}, \Delta \mathrm{G}^{\circ}\) at \(\mathrm{T}_{1}\) and \(\Delta \mathrm{G}^{\circ}\) at \(\mathrm{T}_{2}\) (in \(\mathrm{kj} \mathrm{mol}^{-1}\) ) respectively, are close to [use \(\left.\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]\) (a) \(28.4,-7.14\) and \(-5.71\) (b) \(0.64,-7.14\) and \(-5.71\) (c) \(28.4,-5.71\) and \(-14.29\) (d) \(0.64,-5.71\) and \(-14.29\)

For the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) at a given temperature, the equilibrium amount of \(\mathrm{CO}_{2}(\mathrm{~g})\) can be increased by [1998 - 2 Marks] (a) adding a suitable catalyst (b) adding an inert gas (c) decreasing the volume of the container (d) increasing the amount of \(\mathrm{CO}(\mathrm{g})\).

When equal volumes of the following solutions are mixed, precipitation of \(\mathrm{AgCl}\left(\mathrm{K}_{\mathrm{sp}}=1.8 \times 10^{-10}\right)\) will occur only with (a) \(10^{-4} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-4} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (b) \(10^{-5} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-5} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (c) \(10^{-6} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-6} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (d) \(10^{-10} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-10} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\)
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