In other words, we use the Henderson-Hasselbalch equation (below) to determine the ratio of protonated vs deprotonated molecules in the solution, and therefore the molecules overall charge. [A-] pH = pKa + log [HA] 11. Find the ratio of formate to formic acid when the pK, = 3.7 at a pH of 3.6. 12. Find the pH of a formic acid solution which has a pK, = 3.7 and a ratio of formate to formic acid of 2.3. %3D 13. Find the pH of a formic acid solution when the concentration of formate and formic acid are equal to one another. -- .....

In other words, we use the Henderson-Hasselbalch equation (below) to determine the ratio of protonated vs deprotonated molecules in the solution, and therefore the molecules overall charge. [A-] pH = pKa + log [HA] 11. Find the ratio of formate to formic acid when the pK, = 3.7 at a pH of 3.6. 12. Find the pH of a formic acid solution which has a pK, = 3.7 and a ratio of formate to formic acid of 2.3. %3D 13. Find the pH of a formic acid solution when the concentration of formate and formic acid are equal to one another. -- .....

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Acetic acid (pK, = 4.76) was dissolved in an aqueous solution buffered to a pH of 5.76. Determine the ratio of the concentrations of acetate ion and acetic acid in this solution. Which species predominates under these conditions? O The acid predominates, because the ratio of [acetate ion]/[acetic acid] = 0.01. O Neither species predominates, because the ratio of [acetate ion]/[acetic acid] = 1. O The conjugate base predominates, because the ratio of [acetate ion]/[acetic acid] = 10. The conjugate base predominates, because the ratio of [acetate ion]/[acetic acid] = 100. %3D O The acid predominates, because the ratio of [acetate ion]/[acetic acid] = 0.1.
11.An acid-base equilibrium system is created by dissolving 0.20mol CH3COOH in water and diluting the resulting solution to a volume of 1.0 L. What is the effect of adding 0.020 mol CH3COO (aq) to this solution? How will pH change (calculate pH before and after the addition. Ka of CH3COOH is 1.76 x10-5)? How will concentrations of CH3COOH and CH3COO at equilibrium change?
Write the Henderson-Hasselbalch equation for a propanoic acid solution (CH₂CH₂CO₂H, PK₁ = 4.874). pH || pK₂ + log. Answer Bank [A] [HA]
Ascorbic acid (H₂C,H,O) is a diprotic acid. The acid dissocation constants for H₂C,H,O, are Kal = 8.00 x 10-5 and K₁2 = 1.60 x 10-12 Determine the pH of a 0.117 M solution of ascorbic acid. pH = Determine the equilibrium concentrations of all species in the solution. [H₂C,H,O,] = [HC,H,O,]= [C,H,O2] = M M M
The ionization constant of a very weak acid, HA, is 1.9x10-9. Calculate the equilibrium concentrations of H30", A", and HA in a 0.040 M solution of the acid. Determine the concentrations of all species at equilibrium and the solution pH. [H3O"] =| mol/L [A'] =[ mol L [HA] = | mol L pH =
A student was titrating a solution of hydrazine, H₂NNH₂, with a strong acid solution. Determine the pH at a particular point in the titration. Do this by constructing a BCA table and determining the pH. Complete Parts 1-2 before submitting your answer. NEXT > 40.0 mL of a 0.200 M H₂NNH₂ solution was titrated with 100 mL of 0.100 M HNO3 (a strong acid). Fill in the BCA table with the appropriate value for each involved species to determine the moles of reactant and product after the reaction of the acid and base.. Before (mol) Change (mol) After (mol) 6.00 × 10-³ 0 -6.00 × 10-³ 1 H₂NNH₂(aq) 0.100 7.00 × 10-3 0.200 -7.00 × 10-³ + H+ (aq) 1.00 x 10-² 2 8.00 × 10-³ -1.00 x 10-² -8.00 × 10-³ H₂NNH3 + (aq) 2.00 × 10-³ RESET -2.00 × 10-³ +
+ Base/Acid Ratios in Buffers Just as pH is the negative logarithm of [H3O+]. PK₁ is the negative logarithm of Ka. PK₂ = -log Ka The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions: pH=pK₁ +log base [acid] Notice that the pH of a buffer has a value close to the PK₂ of the acid, differing only by the logarithm of the concentration ratio [base]/[acid]. The Henderson-Hasselbalch equation in terms of pOH and PKb is similar. lacid] pOH = pk + logad base pH 3.74 [acetic acid] ten times greater than [acetate] Submit ✓ Correct ▼ Part B pH = 4.74 Previous Answers [acetate] = [acetic acid] mass of NH4Cl = pH = 5.74 [acetate] ten times greater than [acetic acid] Value How many grams of dry NH4Cl need to be added to 2.00 L of a 0.100 M solution of ammonia, NH3. to prepare a buffer solution that has a pH of 8.85? Kb for ammonia is 1.8 x 10-5 Express your answer with the appropriate units. ▸ View Available Hint(s) ( 2 F Templates Symbols undo relo reset keyboard…
Using the table of the weak base below, you have chosen Pyridine as your weak base in the buffer solution. You have already added enough of the conjugate acid salt to make the buffer solution concentration at 0.62 M in this salt. The desired pH of the buffer should be equal to 4.5. Values of K, for Some Common Weak Bases Name Formula Kb 1.8 x 10-5 Ammonia Methylamine NH3 CHÍNH, 4.38 x 10-4 C₂H5NH₂ 5.6 x 10-4 Ethylamine Aniline CHẠNH, 3.8 x 10-10 Pyridine CsHsN 1.7 x 10-⁹ 2. Compute for pKb Conjugate Acid NH4+ CH;NH * C₂H5NH3+ CH,NH,* CH,NH*
Determine the pH at the point in the titration of 40.0 mL of 0.200 M HC4H,O2 with 0.100 M Sr(OH)2 after 10.0 mL of the strong base has been added. The value of Ka for HC,H,O2 is 1.5 x 10-5. 3 NEXT Use the table below to determine the moles of reactant and product after the reaction of the acid and base You can ignore the amount of liquid water in the reaction. HC4H,O2(aq) OH (aq) H2O(1) C4H,O2 (aq) + Before (mol) 1.00 x 103 8.00 x 103 8.00 x 103 Change (mol) -2.00 x 103 -8.00 x 103 After (mol) 8.00 x 103 6.00 x 103 2.00 x 103 RESET 0.100 0.200 1.00 x 103 -1.00 x 103 2.00 x 103 -2.00 x 103 6.00 x 103 -6.00 x 103 7.00 x 103 -7.00 x 10"3 8.00 x 103 -8.00 x 10"3
a) Why can you not prepare a solution of pH = 8 by diluting sufficiently a 0.01 M solution of HCl with water? b) You could determine Ka for benzoic acid by measuring the pH of a 0.010 M solution of benzoic acid. Why is the graphical method used in this experiment a better way of determining Ka? c) List the measurements along with their uncertainties made in the determination of the molar mass of benzoic acid. Calculate the uncertainty in the mola r mass of benzoic acid determined by the titration.
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of its conjugate acid and the ratio of the concentrations of the conjugate base and acid. The equation is important in laboratory work that makes use of buffered solutions, in industrial processes where pH needs to be controlled, and in medicine, where understanding the Henderson-Hasselbalch equation is critical for the control of blood pH. Part A As a technician in a large pharmaceutical research firm, you need to produce 450. mL of 1.00 M potassium phosphate buffer solution of pH = 6.91. The pK₁ of H₂PO4¯ is 7.21. You have the following supplies: 2.00 L of 1.00 M KH₂PO4 stock solution, 1.50 L of 1.00 M K₂HPO4 stock solution, and a carboy of pure distilled H₂O. How much 1.00 M KH₂PO4 will you need to make this solution? Express your answer to three significant digits with the appropriate units. View Available Hint(s) 2 Volume of KH₂PO4 needed = Value Review | Constants | Periodic Table Submit μÃ Units ?
6. a. Calculate the pH of a buffered solution that is 0.100 M in C6H5CO2H (benzoic acid, Ka = 6.4 x 10-5) and 0.100 M NaC6H5CO2. b. Calculate the pH after 20.0% (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium: C6H5CO2H (aq) + H2O (l) ⇌ C6H5CO2- (aq) + H3O+(aq) c. Do the same as in part b, but use the following equilibrium to calculate the pH: C6H5CO2 -(aq) + H2O(l) ⇌C6H5CO2H (aq) + OH- (aq) d. Do your answers in part b and c agree? Explain.