(HCOOH) = acid = 0.010 M x (100 mL/180 mL) = 0.00555 M
(NaHCOO) = base = 0.01 M x (80 mL/180 mL) = 0.00444 M
pH = pKa + log [(base)/(acid)]===This is the Henderson-Hasselbalch equation.
You will need to look up the pKa for HCOOH or find Ka and convert to pKa as pKa = -logKa. Then plug in the concentrations and solve for pH. This is a normal buffer which is a salt of a weak acid (HCOOH) and a salt of that weak base (NaHCOO). Post your work if you get stuck.
A buffer solution is prepared by combining 100 mL of 0.010 M HCOOH and 80 mL of 0.010 M NaHCOO.
Calculate the pH of this buffer solution and explain the classification of the solution as a buffer.
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1 answer