Answer:
The pH of this solution is 12.1.
Explanation:
1 mole of sulfuric acid gives 2 moles of hydrogen ions, then 0.100 M of sulfuric acid will give :
Volume of sulfuric acid solution = 50.0 mL = 0.050 L ( 1 mL = 0.001 L)
Moles of hydrogen ions in sulfuric acid solution :a
HOCl
The dissociation constant value of HOCl =
initially
0.1133 M 0 0
At equilibrium :
(0.1133-x) x x
The expression of dissociation constant is given as:
Solving for x:
x =
Volume of HOCl solution = 30.0 mL = 0.030 L ( 1 mL = 0.001L)
Moles of hydrogen ions in HOCl solution = b
Total moles of hydrogen ions = a + b
=
1 mole of NaOH gives 1 mole of hydroxide ions, then 0.200 M of NaOH acid will give :
Volume of NaOH solution = 25.0 mL = 0.025 L ( 1 mL = 0.001 L)
Moles of hydroxide ions in NaOH solution : c
1 mole of gives 2 mole of hydroxide ions, then 0.100 M of
Volume of solution = 25.0 mL = 0.025 L ( 1 mL = 0.001 L)
Moles of hydroxide ions in solution : d
1 mole of KOH gives 1 mole of hydroxide ions, then 0.170 M of KOH will give :
Volume of KOH solution = 10.0 mL = 0.010 L ( 1 mL = 0.001 L)
Moles of hydroxide ions in KOH solution : e
Total moles of hydroxide ions : c + d + e
Total moles of hydrogen ions = 0.010002 mol
Total moles of hydroxide ions = 0.0117 mol
1 mole of hydrogen ion reacts with 1 mole of hydroxide ion to from a water with neutral pH.
Hydrogen ions < hydroxide ion
So, 0.0117 moles of hydroxide ion will neutralize 0.0117 moles of hydrogen ions.
Moles of hydroxide left after neutralization = 0.0117 mol - 0.010002 mol
= 0.001698 moles
Concentration of hydroxide ions left in the solution :
pH = 14 - pOH
pH = 14 - 1.9= 12.1
The pH of this solution is 12.1.