Question

When a 22.0 mL sample of a 0.448 M aqueous hydrocyanic acid solution is titrated with a 0.387 M aqueous barium hydroxide solution, what is the pH after 19.1 mL of barium hydroxide have been added? pH =

Answer 1

Balanced Neutralization reaction is

2HCl + Ba(OH)₂    BaCl₂ + 2H₂O

no. of mole = molarity X volume of solution in liter

1 ml = 0.001 liter then

22 ml = 0.001 X 22 = 0.022 liter

19.1 ml = 0.001 X 19.1 = .0191 liter

no. of moles of HCl = 0.448 X 0.022 = 0.009856 moles

no.of moles of Ba(OH)₂ = 0.387 X 0.0191 = 0.0073917 moles

According to balanced reaction 2 mole of HCl react with 1 mole of Ba(OH)₂ molar ratio between HCl to Ba(OH)₂ is 2:1

therefore to react with 0.009856 moles of HCl required Ba(OH)₂ = 0.009856/2 = 0.004928 moles But

Ba(OH)₂ given 0.0073917 moles

excess Ba(OH)₂ remain after neutralization reaction = 0.0073917 - 0.004928 = 0.0024637 moles

Ba(OH)₂ is strong base dissociate as

Ba(OH)₂ + H₂O Ba²⁺_(aq) + 2OH^-_(aq)

thus dissociation of Ba(OH)₂ show that 1 mole Ba(OH)₂ produce 2 mole OH^- therefore 0.0024637 moles of Ba(OH)₂ produce OH^- = 0.0024637 X 2 = 0.0049274 moles

moles of OH^- = 0.0049274 moles

total volume of solution = 0.022 liter 0.0191 liter = 0.0411 liter

molarity = no.of moles / volume of solution in liter

molarity of OH^- = 0.0049274/0.0411 = 0.1199 M

pOH = -log[OH^-] = -log(0.1199) = 0.92

pH = 14 - pOH = 14 - 0.92 = 13.08

pH of solution = 13.08