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Chapter 16: Problem 41

Calculate the concentrations of all the species (HCN, \(\mathrm{H}^{+}, \mathrm{CN}^{-},\) and \(\mathrm{OH}^{-}\) ) in a \(0.15 \mathrm{M} \mathrm{HCN}\) solution.

Short Answer

Expert verified
First, calculate the [H+] and [CN-] using the acid ionization constant equation (Ka), then use the ion product for water (Kw) to find [OH-].

Step by step solution

01

Set up the equilibrium expression for the dissociation of HCN

HCN \( \rightleftharpoons \) H+ + CN- . We write the equilibrium constant expression for this weak acid ionization: Ka = [H+][CN-] / [HCN]. Because initial ion concentrations are zero in a weak acid solution, we can assume pure HCN ionizes slightly to generate x amount of H+ and CN-, while (0.15 - x) M HCN remains. So, the Ka equation becomes: Ka = x^2 / (0.15 - x)
02

Solve for x (concentration of ions)

Given the value of Ka for HCN (6.2 x 10^-10), we can solve the above equation for x. This will give the concentrations of H+ and CN- ions (as both are produced in equal amounts). Due to the small dissociation of weak acids, x will be much smaller than the initial concentration of the acid, so we can simplify the equation to: Ka = x^2 / 0.15 and solve for x. The pH of the solution can then be determined from [H+].
03

Calculate the OH- concentration

Using the ion product for water, Kw = [H+][OH-], and the calculated [H+] in step 2, we can solve for [OH-].

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Most popular questions from this chapter

In a certain experiment a student finds that the \(\mathrm{pHs}\) of \(0.10 M\) solutions of three potassium salts \(K X, K Y\) and \(\mathrm{KZ}\) are 7.0,9.0 , and 11.0 , respectively. Arrange the acids HX, HY, and HZ in order of increasing acid strength.

When chlorine reacts with water, the resulting solution is weakly acidic and reacts with \(\mathrm{AgNO}_{3}\) to give a white precipitate. Write balanced equations to represent these reactions. Explain why manufacturers of household bleaches add bases such as \(\mathrm{NaOH}\) to their products to increase their effectiveness.

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) has the following structure: An oxalic acid solution contains these species in varying concentrations: \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}, \mathrm{HC}_{2} \mathrm{O}_{4}^{-}, \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\), and \(\mathrm{H}^{+}\). (a) Draw Lewis structures of \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\). (b) Which of the four species listed here can act only as acids, which can act only as bases, and which can act as both acids and bases?

\(\mathrm{Al}^{3+}\) is not a Bronsted acid, but \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is Explain.

Give the conjugate acid of each of these bases: (a) \(\mathrm{HS}^{-}\) (b) \(\mathrm{HCO}_{3}^{-}\) (c) \(\mathrm{CO}_{3}^{2-}\) (d) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (e) \(\mathrm{HPO}_{4}^{2-}, \quad\) (f) \(\mathrm{PO}_{4}^{3-}\) (g) \(\mathrm{HSO}_{4}^{-}\) (h) \(\mathrm{SO}_{4}^{2-}\) (i) \(\mathrm{NO}_{2}^{-}\) (j) \(\mathrm{SO}_{3}^{2-}\)
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