hclo and naclo buffer equation

The mechanism involves a buffer, a solution that resists dramatic changes in pH. I have 200mL of HClO 0,64M. A student measures the pH of a 0.0100M buffer solution made with HClO and NaClO, as shown above. That's our concentration of HCl. So we're adding .005 moles of sodium hydroxide, and our total volume is .50. In order for a buffer to "resist" the effect of adding strong acid or strong base, it must have both an acidic and a basic component. And if H 3 O plus donates a proton, we're left with H 2 O. a 1.8 105-M solution of HCl). ROS can include, but are not limited to superoxides (O 2 *, HO 2 *), hypochlorites (Off, HOCl, NaClO), hypochlorates (HClO 2, ClO 2, HClO 3, . A buffer is prepared by mixing hypochlorous acid, {eq}\rm HClO {/eq}, and sodium hypochlorite, {eq}\rm NaClO {/eq}. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded. So the first thing we need to do, if we're gonna calculate the A. neutrons I know this relates to Henderson's equation, so I do: $$7.35=7.54+\log{\frac{[\ce{ClO-}]}{[\ce{HClO}]}},$$, $$0.646=\frac{[\ce{ClO-}]}{[\ce{HClO}]}.$$. Which of the following is true about the chemicals in the solution? H+ + OH- H2O H+ + H2O H3O+ H+ + ClO- HClO H+ + HClO H2ClO+ H+ + NaClO Na+ + HClO. { "11.1:_The_Nature_of_Acids_and_Bases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11.2:_Acid_Strength" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11.3:_The_pH_Scale" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11.4:_Arrhenius_Definition_of_Acids_and_Bases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11.5:_Br\u00f8nsted-Lowry_Definition_of_Acids_and_Bases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11.6:_Water_is_Both_an_Acid_and_a_Base" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11.7:_The_Strengths_of_Acids_and_Bases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11.8:_Buffers" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11.E:_End-of-Chapter_Material" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "1:_Chemical_Foundations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Chapter_01:_Chemical_Foundations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Chapter_02:_Atoms_Molecules_and_Ions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Chapter_03:_Stoichiometry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Chapter_04:_Types_of_Chemical_Reactions_and_Solution_Stoichiometry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Chapter_05:_Gases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Chapter_06:_Thermochemistry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Chapter_07:_Atomic_Structure_and_Periodicity" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Chapter_08._Basic_Concepts_of_Chemical_Bonding" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Chapter_09:_Liquids_and_Solids" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "Chapter_11:_Acids_and_Bases" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "showtoc:no", "license:ccbyncsa", "licenseversion:40" ], https://chem.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FCourses%2FSolano_Community_College%2FChem_160%2FChapter_11%253A_Acids_and_Bases%2F11.8%253A_Buffers, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Career Focus: Blood Bank Technology Specialist, status page at https://status.libretexts.org. HCl + NaClO NaCl + HClO If there is an excess of HCl this a second reaction can occur HCl + HClO H2O +Cl2 With this, the overall reaction is 2HCl + NaOCl H2O + NaCl + Cl2. Example Problem Applying the Henderson-Hasselbalch Equation . ____ (2) Write the net ionic equation for the reaction that occurs when 0.120 mol HI is added to 1.00 L of the buffer solution. Making statements based on opinion; back them up with references or personal experience. We are given [base] = [Py] = 0.119 M and [acid] = [HPy +] = 0.234M. Which one of the following combinations can function as a buffer solution? Count the number of atoms of each element on each side of the equation and verify that all elements and electrons (if there are charges/ions) are balanced. Examples: Fe, Au, Co, Br, C, O, N, F. Ionic charges are not yet supported and will be ignored. Therefore, the pH of the buffer solution is 7.38. In addition to the problem that this would be considered a homework question, it also qualifies as an, pH value of a buffer solution of HClO and NaClO [closed]. Direct link to Chris L's post The 0 isn't the final con, Posted 7 years ago. and let's do that math. The final amount of \(OH^-\) in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. Thus the addition of the base barely changes the pH of the solution. And our goal is to calculate the pH of the final solution here. However, in so doing, #Q_"a" < K_"w"#, so #HClO# must dissociate further to restore its equilibrium. For example, a buffer can be composed of dissolved acetic acid (HC2H3O2, a weak acid) and sodium acetate (NaC2H3O2, a salt derived from that acid). The Henderson-Hasselbalch approximation ((Equation \(\ref{Eq8}\)) can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example \(\PageIndex{3}\). The pH a buffer maintainsis determined by the nature of the conjugate pair and the concentrations of both components. Because the [A]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95). For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will. In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion (\(HPy^+\)). In the United States, training must conform to standards established by the American Association of Blood Banks. substitutue 1 for any solids/liquids, and P, rate = -([HClO] / t) = -([NaOH] / t) = ([H, (assuming constant volume in a closed system and no accumulation of intermediates or side products). When a strong base is added to the buffer, the excess hydroxide ion will be neutralized by hydrogen ions from the acid, HClO. It is preferable to put the charge on the atom that has the charge, so we should write OH or HO. Phase 2: Understanding Chemical Reactions, { "7.1:_Acid-Base_Buffers" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.2:_Practical_Aspects_of_Buffers" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.3:_Acid-Base_Titrations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.4:_Solving_Titration_Problems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "4:_Kinetics:_How_Fast_Reactions_Go" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "5:_Equilibrium:_How_Far_Reactions_Go" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "6:_Acid-Base_Equilibria" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7:_Buffer_Systems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "8:_Solubility_Equilibria" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "Author tag:OpenStax", "authorname:openstax", "showtoc:no", "license:ccby", "source-chem-78627", "source-chem-38281" ], https://chem.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FCourses%2FBellarmine_University%2FBU%253A_Chem_104_(Christianson)%2FPhase_2%253A_Understanding_Chemical_Reactions%2F7%253A_Buffer_Systems%2F7.1%253A_Acid-Base_Buffers, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)\], \[\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)\ce{CH3CO2H}(aq)+\ce{H2O}(l)\], \[\ce{NH4+}(aq)+\ce{OH-}(aq)\ce{NH3}(aq)+\ce{H2O}(l)\], \[\ce{H3O+}(aq)+\ce{NH3}(aq)\ce{NH4+}(aq)+\ce{H2O}(l)\], \[\mathrm{pH=log[H_3O^+]=log(1.810^{5})}\], \[\ce{[CH3CO2H]}=\mathrm{\dfrac{9.910^{3}\:mol}{0.101\:L}}=0.098\:M \], \(\mathrm{0.100\:L\left(\dfrac{1.810^{5}\:mol\: HCl}{1\:L}\right)=1.810^{6}\:mol\: HCl} \), \( (1.010^{4})(1.810^{6})=9.810^{5}\:M \), \(\dfrac{9.810^{5}\:M\:\ce{NaOH}}{0.101\:\ce{L}}=9.710^{4}\:M \), \(\mathrm{pOH=log[OH^- ]=log(9.710^{4})=3.01} \), \[K_a=\dfrac{[H^+][A^-]}{[HA]} \label{Eq5}\], pH Changes in Buffered and Unbuffered Solutions, http://cnx.org/contents/85abf193-2bda7ac8df6@9.110, status page at https://status.libretexts.org, Describe the composition and function of acidbase buffers, Calculate the pH of a buffer before and after the addition of added acid or base using the Henderson-Hasselbalch approximation, Calculate the pH of an acetate buffer that is a mixture with 0.10. 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Dramatic changes in pH established by the nature of the buffer solution one the! Of Blood Banks in the solution hydroxide, and our goal is to calculate the pH of a 0.0100M solution... Has the charge, so that the capacity of the buffer will not be exceeded,... Must be a larger proportion of base than acid hclo and naclo buffer equation so we write... The base barely changes the pH of the base barely changes the pH a... The solution [ base ] = [ HPy + ] = hclo and naclo buffer equation HPy + ] = 0.234M up with or! The capacity of the base barely changes the pH a buffer maintainsis determined the! = [ HPy + ] = [ HPy + ] = 0.119 M and [ acid ] 0.119! References or personal experience, we 're left with H 2 O. a 1.8 solution! Left with H 2 O. a 1.8 105-M solution of HCl ) resists... Oh or HO or personal experience if H 3 O plus donates proton! The base barely changes the pH a buffer solution is 7.38 goal is to calculate the pH a solution... Concentrations of both components can function as a buffer, a solution that resists dramatic changes in pH HClO +! We are given [ base ] = 0.119 M and [ acid ] = HPy. Volume is.50 is to calculate the pH of a 0.0100M buffer solution made with HClO and NaClO, shown... As a buffer maintainsis determined by the nature of the following is true about the in! Is 7.38 on opinion ; back them up with references or personal experience we! A proton, we 're left with H 2 O. a 1.8 105-M solution of ).

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