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Chapter 14: Problem 78

A \(1.0 \times 10^{-2}-M\) solution of cyanic acid (HOCN) is 17\(\%\) dissociated. Calculate \(K_{\mathrm{a}}\) for cyanic acid.

Short Answer

Expert verified
The Ka for cyanic acid can be calculated using the equilibrium concentrations of the dissociation products and the undissociated acid: \(K_a = \frac{(0.0017)(0.0017)}{0.0083} \approx 3.48 \times 10^{-4}\).

Step by step solution

01

Write the dissociation equation for the acid

Write the dissociation equation for cyanic acid (HOCN) in water: \(HOCN \longleftrightarrow H^+ + OCN^-\)
02

Determine the initial concentrations of HOCN, H+, and OCN-

We are given a 0.01 M solution of HOCN. Both the H+ and OCN- ions will be formed as the dissociation progresses, so their initial concentrations will be 0. Initial concentrations: [HOCN] = 0.01 M [H+] = 0 M [OCN-] = 0 M
03

Determine the equilibrium concentrations of HOCN, H+, and OCN-

Since 17% of the HOCN molecules have dissociated, we can determine the equilibrium concentrations by considering a decrease of 0.17 * 0.01 M for HOCN and an increase of the same amount for H+ and OCN-: Equilibrium concentrations: [HOCN] = 0.01 M - 0.17 * 0.01 M [H+] = 0 + 0.17 * 0.01 M [OCN-] = 0 + 0.17 * 0.01 M Calculate the values: [HOCN] = 0.01 M - (0.17 * 0.01 M) = 0.0083 M [H+] = 0.17 * 0.01 M = 0.0017 M [OCN-] = 0.17 * 0.01 M = 0.0017 M
04

Write the equilibrium constant expression (Ka) and calculate Ka

Now that we have the equilibrium concentrations of all species, we can write the expression for Ka: \(K_a = \frac{[H^+][OCN^-]}{[HOCN]}\) Plug in the equilibrium concentrations: \(K_a = \frac{(0.0017)(0.0017)}{0.0083}\) Calculate Ka: \(K_a \approx 3.48 \times 10^{-4}\) So the Ka for cyanic acid is approximately \(3.48 \times 10^{-4}\).

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Most popular questions from this chapter

An acid \(\mathrm{HX}\) is 25\(\%\) dissociated in water. If the equilibrium concentration of \(\mathrm{HX}\) is \(0.30 \mathrm{M},\) calculate the \(K_{\mathrm{a}}\) value for \(\mathrm{HX}\) .

A typical sample of vinegar has a pH of \(3.0 .\) Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{\mathrm{a}}=1.8 \times\right.\) \(10^{-5}\) ), calculate the concentration of acetic acid in vinegar.

Would you expect \(\mathrm{Fe}^{3+}\) or \(\mathrm{Fe}^{2+}\) to be the stronger Lewis acid? Explain.

Calculate the pH of the following solutions: a. 1.2\(M \mathrm{CaBr}_{2}\) b. 0.84$M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}} \text { for } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\right)$ c. 0.57$M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)$

Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced equations for the reactions causing the solution to be acidic or basic. The relevant \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values are found in Tables 14.2 and \(14.3 .\) $\begin{array}{ll}{\text { a. } \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}} & {\text { d. } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{ClO}_{2}} \\ {\text { b. } \mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}} & {\text { e. } \mathrm{NH}_{4} \mathrm{F}} \\ {\text { c. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{O} \mathrm{l}} & {\text { f. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{CN}}\end{array}$
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