Base/Acid Ratios in Buffers Acetic acid has a Ka of 1.8 × 10-5 . Three acetic ac

Question

Base/Acid Ratios in Buffers Acetic acid has a Ka of 1.8 × 10-5 . Three acetic acid/acetate buffer solutions. A, B, and C, were made using varying concentrations: Just as pH is the negative logarithm of [HO pKa is the negative logarithm of K 1. acetic acid ten times greater than acetate 2. acetate ten times greater than acetic acid. and 3. acetate] =[acetic acid The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions: Match each buffer to the expected pH. Drag each item to the appropriate bin. pH= pK, + log-base Hints Notice that the pII of a buffer has a value close to the pKa of the acid, differing only by the logarithm of the concentration ratio [base / acid] Reset Help pH = 3.74 pH = 4.74 pH 5.74 Submit My Answers Give Up Correct Part B How many grams of dry NHCl need to be added to 2.20 L of a 0.600 M solution of ammonia, NHs, to prepare a buffer solution that has a pH of 8.60? K for ammonia is 1.8 x 10-

Explanation / Answer

Solution- pOH = 14 .- 8.68 = 5.32
pKb = 4.74

5.32 - 4.74 = 0.58

10^0.58 = 3.80 = [NH4+]/ 0.600

[NH4+] = 0.600 x 3.80=2.28 M

moles NH4+ = 2.28 x 2.20 L

=5.016

moles NH4Cl = 5.016 x 53.492 g/mol=268.31 g

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