When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. Buffers resist change in ( mathrm{pH} ) following the addition of acid or base. A buffer solution prepared from a weak acid (HA) and its conjugate base ( left(A^{-} ight) )is represented as [ mathrm{HA}(mathrm{aq}) ightleftharpoons mathrm{H}^{+}(mathrm{aq})+mathrm{A}^{-}(mathrm{aq}) ] The buffer will follow Le Châtelier's principle. If acid is added, the reaction shifts to consume the added ( mathrm{H}^{+} ), forming more HA. When base is added, the base will react with ( mathrm{H}^{+} ), reducing its concentration. The reaction then shifts to replace ( mathrm{H}^{+} )through the dissociation of ( mathrm{HA} ) into ( mathrm{H}^{+} )and ( mathrm{A}^{-} ). In both instances, ( left[mathrm{H}^{+} ight] )tends to remain constant. The ( mathrm{pH} ) of a buffer is calculated by using the Henderson-Hasselbalch equation: [ mathrm{pH}=mathrm{p} K_{mathrm{a}}+log frac{left[mathrm{A}^{-} ight]}{[mathrm{HA}]} ] Part A What is the pH of a buffer prepared by adding ( 0.809 mathrm{~mol} ) of the weak acid ( mathrm{HA} ) to ( 0.305 mathrm{~mol} ) of ( mathrm{NaA} ) in ( 2.00 mathrm{~L} ) of solution? The dissociation constant ( K_{mathrm{a}} ) of ( mathrm{HA} ) is ( 5.66 imes 10^{-7} ). Express the ( mathrm{pH} ) numerically to three decimal places. View Available Hint(s) ( mathrm{pH}=5.823 ) Submit Previous Answers Correct Since both the acid and base exist in the same volume, we can skip the concentration calculations and use the number of moles in the Henderson-Hasselbalch equation to calculate the ( mathrm{pH} ). The answer will be the same. Part B What is the ( mathrm{pH} ) after ( 0.150 mathrm{~mol} ) of ( mathrm{HCl} ) is added to the buffer from Part ( mathrm{A} ) ? Assume no volume change on the addition of the acid. Express the ( mathrm{pH} ) numerically to three decimal places. View Available Hint(s)
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When HCl is added to the buffer, it will react with the base (A-) in the buffer. This is because HCl is a strong acid and will completely dissociate into H+ and Cl-. The H+ ions will react with the base (A-) to form the weak acid (HA). This means that the amount Show more…
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When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. Buffers resist changes in pH following the addition of acid or base. A buffer solution prepared from a weak acid (HA) and its conjugate base (A-) is represented as HA(aq) ⇌ H+(aq) + A-(aq). The buffer will follow Le Châtelier's principle. If acid is added, the reaction shifts to consume the added H+, forming more HA. When base is added, the base will react with H+, reducing its concentration. The reaction then shifts to replace H+ through the dissociation of HA into H+ and A-. In both instances, [H+] tends to remain constant. The pH of a buffer is calculated using the Henderson-Hasselbalch equation: pH = pKa + log[A-]/[HA]. a) What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.609 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10^-7. b) What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. c) What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.
Adi S.
Henderson-Hasselbalch Equation For each conjugate acid/base pair in a buffer system, the relationship between pH and the ratio of concentrations of the weak acid and its conjugate base is given by the Henderson-Hasselbalch equation: You are asked to make a 50 mL of a 0.1 M phosphate buffer with a pH of 7.4. From the previous question, you determined the ratio of [A-] to [HA] (or the buffer strength). Using the buffer strength you calculated in the previous question, what is the concentration (in M) of the base, [A-], and of the acid, [HA] in this buffer? where pH = -log10[H+], pKa = -log10(Ka), and concentrations are in units of molarity (M). If a strong acid or a strong base is added to a buffer system, NEW values of [HA] and [A-] need to be calculated before being used in the Henderson-Hasselbalch equation. To do this, first determine the initial moles of HA and A- from the initial concentrations. Then, determine the moles of the strong acid (or strong base) that has been added. Since a strong acid will be neutralized by the base component of the buffer, the moles of strong acid will be subtracted from the initial moles of the conjugate base (A-). When this neutralization occurs, more acid (HA) is produced so the moles of strong acid will be added to the initial moles of the weak acid (HA). With these new mole values, divide each by the new total volume and now you have concentrations that can be used in the Henderson-Hasselbalch equation. Hint: Recall that [HA] + [A-] = [buffer] Express each answer to three places past the decimal. [A-] = M and [HA] = M
b) i) The first buffer is 100.0 mL of a buffer that is 0.25 M in formic acid (HCO2H, Ka = 1.8E-4) and 0.15 M in sodium formate, NaCO2H. If a 10.0 mL sample of 0.10 M NaOH is added to this buffer, what will be the pH? ii) What is the increase in pH? c) Write the common ion formulas for the following: HNO2 and KNO2, NH3 and NH4Cl, C2H5NH2 and C2H5NH3Br. d) In Question 7, you were asked to calculate the pH of a buffer solution that is 0.50 M in HClO and 0.40 M in NaClO. What would be the pH of a buffer solution that is 0.40 M in HClO and 0.20 M in NaClO? Ka = 2.9x10^-8. e) Which is the correct net ionic equation when a strong base is added to the buffer made with HCO3- and CO32-? Question options: 1. OH-(aq) + HCO3-(aq) → H2O(l) + CO32-(aq) 2. H2O(l) + HCO3-(aq) ⇌ H2CO3(aq) + OH-(aq) 3. Two of the choices 4. None of the choices 5. H2O(l) + CO32-(aq) ⇌ HCO3-(aq) + OH-(aq) f) A 10.0 mL sample of 0.10 M HCl is added to 100.0 mL of a buffer that is 0.25 M in formic acid (HCO2H, Ka = 1.8E-4) and 0.15 M in sodium formate, NaCO2H. What would the NET IONIC equation be if NaOH were added to the buffer instead of the HCl? Question options: 1. HCO2H + KOH → HCO2H + H2O 2. HCO2H + OH- → CO2H- + H2O 3. HCO2H + CO2H- + KOH → KCO2H + H2O + OH- 4. CO2H- + OH- → HCO2H + H2O
Ashton S.
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