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“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”
AqueousEquilibria
What is the PH of a solution made by adding 0.3 mol of acetic acid and 0.3 mol of soduim acetate to enough water to make 1L of solution.
A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.
*What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOKaq (0.52 M ) → K+
aq + HCOO-aq ( 0.52 M )
HCOOH ↔ H+aq + HCOO-
aq
initial : 0.30M 0.00 0.52M Change : -x +x +x Equilibrium: 0.30-x x 0.52+x , 0.30-x ≈ x , 052+x ≈ 0.52 pH = pKa + log [HCOO- ] / [HCOOH] = 3.77 + log [0.52] / [0.30] = 4.01 ----------------------------------------------------------------------------------------------------- * Which of the following are buffer system ? a) KF /HF (buff.) b) KBr/HBr ( not buff.) c) Na2CO3 /NaHCO3 (buff.) d) KH2PO4 / H3PO4 ( buff.) e) NaClO4 / HClO4 (not buff.) f) C5H5N/C5H5NH+ (buff. ) g) K2HPO4 / H3PO4 ( not buff.) .
----------------------------------------------------------Calculate the pH of the 0.30 M NH3 /0.36 M NH4Cl buffer system. What is the pHAfter the addition of 0.01 mol of NaOH to 1.0 L of buffer solution ? . Assume thatthe volume of the solution does not change when the NaOH is added . NH4
+ (aq) ↔ H+
(aq) + NH3 (aq) , pH = pKa + log [NH3] /[NH4 +]
pH = 9.25 + log [0.30] / [0.36] = 9.17 : NH4
+(aq) + OHaq → H2O(l) + NH3(aq)
start : 0.36 M 0.01 M 0.30 M end : 0.35 M 0.00 0.31 M pH = 9.25 + log [0.31] / [0.35] = 9.20 --------------------------------------------------------------------------------------------------------------