Question

1) If one liter of nitric acid solution (pH = 1.000) is mixed with nine liters of barium
hydroxide solution (pH = 11.000), what will be the pH of the resulting solution?

2) The pH of a 0.10M solution of barium acetate is:_______

Answer 1

1 L of HNO3 , pH = 1.00

[H⁺] = 10^-1

[HNO₃] = 0.1 M or 0.1 mol / 1000 mL

Ba(OH)₂ , pH = 11.00

pOH = 14 - 11 = 3.0

[OH^-] = 10^-3

[OH^-] = 0.001 M or 0.001 mol/1000 mL

1 mol of Ba(OH)₂ contains 2 mol of [OH^-]

So0 concentration of Ba(OH)₂ = 0.0005 M   or 0.0005 mol / 1000 mL

Total moles of Ba(OH)₂ = (0.0005 mol / 1 L) * 9 L => 0.0045 mol of Ba(OH)₂

And total moles of [OH^-] = 0.009 mol

After neutralization H⁺ + OH^-   ---> H₂O

moles of [H⁺] = 0.1 - 0.009 => 0.091 moles

Total volume = 1 + 9 = 10 L

[H⁺] = 0.091 moles / 10 L => 0.0091

pH = - log ( 0.0091)

pH = 2.04

2)

Ba(OAc)₂ + 2 H₂O ----> 2 AcOH + Ba(OH)₂

pH of hydrolized salt composed by strong base and weak acid can be calculated as

pH = 7 + 1/2 pKa + 1/2 log C

pKa of acetic acid = 4.76

C = concentration of salt

C = 2*0.1 = 0.2 M ( after dissociation into water Ba(OAc)₂ , it produces 2 eq oa weak acid and 2 eq of OH- ion )

pH = 7 + 1/2 *4.76 + 1/2 log(0.2)

pH = 7 + 2.38 + ( -0.35)

pH = 9.03