Please PLease help! Just as pH is the negative logarithm of [H3O+], pKa is the n
ID: 827724 • Letter: P
Question
Please PLease help!
Just as pH is the negative logarithm of [H3O+], pKa is the negative logarithm of Ka, pKA = - log KA The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions: PH = PKa+log[base]/[acid] Notice that the pH of a buffer has a value close to the pjRTa 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. pOH = pKb + log [acid]/[base] How many grams of dry NH4CI need to be added to 2.00L of a 0.300M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.93? Kb for ammonia is 1.8 times 10-5.Explanation / Answer
pOH = 14 - 8.93 = 5.07
pKb = 4.74
5.07 - 4.74 = 0.33
10^0.33 = 2.137 = [NH4+]/ 0.300
[NH4+] = 0.300 x 2.137=0.641 M
moles NH4+ = 0.641 x 2.00 L=1.282
moles NH4Cl = 1.282 x 53.492 g/mol=68.57 g