A solution made by mixing 95.00 mL of 0.950 M HCOOH with 35.00 mL of 0.750 M NaOH.
3 answers
what about it?
Using tabulated Ka and Kb values, calculate the pH of the following solutions.
millimoles HCOOH = mL x M = 95.00 mL x 0.950 M = 90.25
millimols NaOH = 35.00 mL x 0.750 = 26.25
................HCOOOH + NaOH ==> HCOONa + H2O
Initial............90.25...........0.................0................
add..............................26.25........................................
change.........-26.25.....026.25............+26,25
equilibrium.... 64.00.........0...................26.25
Total volume = 95.00 + 35.00 = 130 mL.
This should be obvious you have a buffered solution of HCOOH and NaOOCH
From the equilibrium line you have
(HCOOH) = millimoles/mL = ? = acid
(HCOONa) = millimoles/mL = ? = base
Substitute into Henserson-Hasselbalch equation which is
pH = pKa + log [(base)/(acid)]
millimols NaOH = 35.00 mL x 0.750 = 26.25
................HCOOOH + NaOH ==> HCOONa + H2O
Initial............90.25...........0.................0................
add..............................26.25........................................
change.........-26.25.....026.25............+26,25
equilibrium.... 64.00.........0...................26.25
Total volume = 95.00 + 35.00 = 130 mL.
This should be obvious you have a buffered solution of HCOOH and NaOOCH
From the equilibrium line you have
(HCOOH) = millimoles/mL = ? = acid
(HCOONa) = millimoles/mL = ? = base
Substitute into Henserson-Hasselbalch equation which is
pH = pKa + log [(base)/(acid)]
1 answer