Can someone solve this for me?
10.0mL of 0.10M HCl is added to a 100.0 mL of a 0.10M solution of NaCO2H (sodium formate).
What is the pH? (Ka for HCO2H = 1.8 x10-4)
This is a buffered solution and you solve these by using the Henderson-Hasselbach equation of pH = pKa + log [(base)/(acid)]. NOTE: To be technically correct one SHOULD substitute concentrations for base and acid; however, since the volume is the same one may use mols. I think it's easier and the answer comes out the same. Additionally I have used millimoles instead of mols BUT I can get away with that because both numerator and denominator were multiplied by 1000 so that cancels also. Using millimoles I don't have to keep with all of those zeros in front of the decimal. 0.001 vs 1.0 and 0.01 vs 10
millimoles HCl = 10 mL x 0.1 M = 1.0
millimoles NaHCOO = 100 mL x 0.1M = 10
................HCOO^^- + HCl ==> HCOOH + Cl^-I
I..................10...............0..............0.............0
add................................1.0...................................
C...................-1.0..........- 1.0.........+ 1.0.........................
E...................9.0.................0..........1.0........................
Substitute into HH equation and solve for pH.
HCOOH is the acid. HCOO^- is the base. Post your work if you would like for me to check it.