Consider a solution formed by mixing 50.0 mL of 0.100 M H2SO4, 30.0 mL of 0.1133 M HOCl, 25.0 mL of 0.200 M NaOH, 25.0 mL of 0.100 M Ba(OH)2, and 10.0 mL of 0.170 M KOH. Calculate the pH of this solution. Ka(HOCl) = 3.5×10-8 pH =

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.

i'm assuming that the correct answer is indeed the first one/ "a"..