Let's check the phone application "buffers" for it's ability to calculate how to prepare a phosphate buffer. Suppose I want 1 liter of a 0.1 M phosphate buffer at a pH of 6.8. Phosphate has a pK2 of 7.2 MW of monosodium phosphate NaH2PO4 (aka monobasic) is 119.98 MW of disodium phosphate Na2HPO4 (aka dibasic) is 141.96 Let's first calculate the fraction of dibasic we need from the H-H equation. In other words solve for A-/total. Calculate to 3 sig figures

Given the values of K and Kb below, which of the systems given would be most suitable to use to construct a buffer of pH 97 HCOOH, K, = 1.8 x 104 CH3COOH, K3=18x105 H3ASO4. Ka1 = 6.0 x 10, Ka2 = 1.1 x 10, Ka3 = 3 x 1012 C5H5N, Kp = 1,7x10-9 NH3 КЬ - 1.3 х. 10-5 OA H3ASO4/NaH2As04 eB. HCOOH/NAHCOO OC. C5H5N/ C5H5NHCI OD.CH3COOH/KCH3CO0 FOE NH4C/NH3

A different titration experiment using a 0.122M standardized NaOH solution to titrate a 26.48 mL solution with an unknown Molarity concentration (M) of sulfuric acid (H2SO4) gave the following molarities for 3 trials.    Initial Burette Reading (mL) Final Burette Reading (mL) Delivered vol (mL) Acid Concentration (M) Trial 1       0.0345 Trial 2       0.0334 Trial 3       0.0381 From the 3 trials, determine the average Molarity concentration of the H2SO4 to 3 significant digits. Don't include a unit.

Calculate the pH of a dihydrogen phosphate/hydrogen phosphate buffer prepared with each of the following concentrations: 1. 0.10 M NaH2РОд and 0.10 Mм NazHPO4 2. 1.0 M NaH2РOд and 1.0 M NazHPO4 3. 0.50 M NaH2PO4 and 0.50M Na2HPO4 What do you conclude about the pH of a buffer when equal concentrations of the weak acid and conjugate base are used to prepare it?

You have 0.100 M solutions of acetic acid (pK₁ = 4.76) and sodium acetate. If you wanted to prepare 1.00 L of 0.100 M acetate buffer of pH 4.00, how many milliliters of acetic acid and sodium acetate would you add? acetic acid: mL sodium acetate: mL

+ Base/Acid Ratios in Buffers Just as pH is the negative logarithm of [H3O+]. PK₂ is the negative logarithm of Ka. PK₂ -log Ka The Henderson-Hasselbalch equation is used buffer solutions: calculate the pH of base DAGSK pH =pK₂ +log 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. acid [base pOH = pkt + log- Part A Acetic acid has a K₁ of 1.8 x 105. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: A. [acetic acid] ten times greater than [acetate]. B. [acetate] ten times greater than [acetic acid], and c. [acetate] [acetic acid]. Match each buffer to the expected pH Drag the appropriate items to their respective bins. ▸ View Available Hint(s) [acetic acid] ten times greater than [acetate] pH = 3.74 [acetate] ten times greater than [acetic acid] pH = 4.74 Reset Help…

You prepare a buffer solution by dissolving 2.00 g each of benzoic acid (HC7H5O2, MW = 122.12) and sodium benzoate (NaC7H5O2, MW = 144.10) in 750.0 mL of water.  The pKb of sodium benzoate is 9.80.  What is the pH of this buffer solution?  [Assume no volume change.]

A buffer solution is made using a weak acid, HA, with a pK₁ of 4.57. If the ratio of [A] to [HA] is 1.0 x 103, what is the pH of the buffer? pH = MacBook Pro

If the pKa of HCHO2 is 3.74 and the pH of an HCHO2/NACHO2 solution is 4.15, which of the following is TRUE? OIt is not possible to make a buffer of this pH from HCHO2 and NaCHO2. O (NACHO2) (NaCHO2) O HCHO2) < [NACHO2)

The pK₂ value for HCN is 9.40. Would a buffer prepared from HCN and NaCN with a pH of 7.60 be considered to be an effective buffer? A buffer in which the mole ratio of NaCN to HCN is 2.0 has a pH of 9.71. Would this buffer solution have a greater capacity for added acid ( H3O+) or added base (OH)?