Sorted by: 11. Determine the acid dissociation constant for a 0.010 M nitrous acid solution that has a pH of 2.70. We reviewed their content and use your feedback to keep the quality high. Step 3: Write the equilibrium expression of Ka for the reaction. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water. A strong base yields 100% (or very nearly so) of OH and HB+ when it reacts with water; Figure \(\PageIndex{1}\) lists several strong bases. It is represented as {eq}pH = -Log[H_{3}O]^+ {/eq}, The pH equation can also be algebraically re-written to solve for the concentration of hydronium ions: {eq}\left [ H_{3}O \right ]^{+} = 10^{-pH} {/eq}, Ka: is the acid disassociation constant and measures how well an acid dissociates in the solution, such as in water. What is the value of Kb for caffeine if a solution at equilibrium has [C8H10N4O2] = 0.050 M, \(\ce{[C8H10N4O2H+]}\) = 5.0 103 M, and [OH] = 2.5 103 M? Write the acid dissociation equation for the dissociation of the weak acid H_2PO_4^- in water. Calculate the present dissociation for this acid. $$\ce{H2SO4 <=> H+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$, $$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$. An acid has a pKa of 6.0. Calculate the pH of 0.060 M HNO2. Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. How does the Hammett acidity function work and how to calculate it for [H2SO4] = 1,830? WebWhen HNO2 is dissolved in water, it partially dissociates according to the equation HNO2H+ + NO2- . (Ka = 4.5 x 10-4), 1. Step 2: Create an Initial Change Equilibrium (ICE) Table for the disassociation of the weak acid. A solution contains 7.050 g of HNO2 in 1.000 kg of water. d. HCN (hydrocyanic acid). Chlorous acid. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure \(\PageIndex{4}\)). HNO_2 iii. What is the pH of a 0.085 M solution of nitrous acid (HNO_2) that has a K_a of 4.5 times 10^{-4}? We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of \(\ce{HSO4-}\) so that we can use \(\ce{[H3O+]}\) to determine the pH. b. The aq stands for aqueous something that is dissolved in water.CH3COOH is a weak acid so only some of the H atoms will dissociate. Write the acid-dissociation reaction of nitrous acid We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table: \[\begin{align*} \ce{[H3O+]} &=~0+x=0+9.810^{3}\:M. \\[4pt] &=9.810^{3}\:M \end{align*} \nonumber \]. 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The equilibrium constant for an acid is called the acid-ionization constant, Ka. This gives: \[K_\ce{a}=1.810^{4}=\dfrac{x^{2}}{0.534} \nonumber \], \[\begin{align*} x^2 &=0.534(1.810^{4}) \\[4pt] &=9.610^{5} \\[4pt] x &=\sqrt{9.610^{5}} \\[4pt] &=9.810^{3} \end{align*} \nonumber \]. Multiplying the mass-action expressions together and cancelling common terms, we see that: \[K_\ce{a}K_\ce{b}=\ce{\dfrac{[H3O+][A- ]}{[HA]}\dfrac{[HA][OH- ]}{[A- ]}}=\ce{[H3O+][OH- ]}=K_\ce{w} \nonumber \]. What is the pH of a 0.0205 M aqueous solution of nitrous acid, HNO2? {eq}Ka = \frac{\left [ H_{3}O^{+}\right ]\left [CH_{3}COO^{-} \right ]}{\left [ CH_{3}COOH \right ]} {/eq}, Step 4: Using the given pH, solve for the concentration of hydronium ions present with the formula: {eq}\left [ H_{3}O \right ]^{+} = 10^{-pH} {/eq}, {eq}\left [ H_{3}O \right ]^{+} = 10^{-2.52} {/eq}, {eq}\left [ H_{3}O \right ]^{+} = 0.003019 M {/eq}. )%2F16%253A_AcidBase_Equilibria%2F16.06%253A_Weak_Acids, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Calculation of Percent Ionization from pH, Equilibrium Concentrations in a Solution of a Weak Acid, Equilibrium Concentrations in a Solution of a Weak Base. Calculate the pH of a 0.150 M solution of nitrous acid, HNO2, pKa = 3.35, assuming that you can neglect the dissociation of the acid in calculating the remaining [HNO2]. Nitrous acid is a weak monoprotic acid and the equilibrium equation of interest is HNO2 + H2O <-> H3O+ + NO2-. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. {/eq} and its acidity constant expression. Thanks for contributing an answer to Chemistry Stack Exchange! By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. That is, when \dfrac{\begin{bmatrix}H_3O^+\end{bmatrix{\begin{bmatrix}c_0\end{bmatrix = \dfrac{1}{2}, Calculate the pH of a solution that is 0.322 M nitrous acid (HNO2) and 0.178 M potassium nitrite (KNO2). Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Ka = 4.5 x 10-4 1. Only a small fraction of a weak acid ionizes in aqueous solution. \[K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=1.8 \times 10^{5} \nonumber \]. Recall that the percent ionization is the fraction of acetic acid that is ionized 100, or \(\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]_{initial}}}100\). \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\), \(K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}\), \(K_a \times K_b = 1.0 \times 10^{14} = K_w \,(\text{at room temperature})\), \(\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}100\). ionic equations - CHEMISTRY COMMUNITY What is the pH of a solution that is 0.50 M in CH3NH3Cl? Using the relation introduced in the previous section of this chapter: \[\mathrm{pH + pOH=p\mathit{K}_w=14.00}\nonumber \], \[\mathrm{pH=14.00pOH=14.002.37=11.60} \nonumber \]. b. HClO_2 (chlorous acid). So we're gonna plug that into our Henderson Formulate an equation for the ionization of the depicted acid. Likewise, for group 16, the order of increasing acid strength is H2O < H2S < H2Se < H2Te. Calculate the pH of a 0.97 M solution of carbonic acid. WebThe chemical equation for the dissociation of HNO2 in water is: HNO2 (aq) H+ (aq) + NO2- (aq)What are the equilibrium concentrations of HNO2 (aq) and NO2- (aq) and the A large Ka value indicates a stronger acid (more of the acid dissociates) and small Ka value indicates a weaker acid (less of the acid dissociates). Fill in the missing value in the following equation: (4.6x10^-4) = (?/HNO2). (a) 2.21 (b) 5.33 (c) 3.35 (d) 4.42. Which of the following options correctly describe the effect of adding solid KClO2 to this system? (Ka = 4.5 x 10-4).
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