Answer to Question #196256 in Chemistry for Kenan

Question

Calculate the concentration of OH ions when reacting 150 cm3 of a solution of a dihydroxylic base, concentration 0.1 mol / dm3 with 90 cm3 of a solution of a troprotic acid with a concentration of 0.1 mol / dm3!

Answer 1

The number of moles of dihydroxylic base in the solution:

n(base) = M(base) × V(base) = 0.1 mol/dm³ × 150 cm³ = 0.1 mol × 0.150 dm³ = 0.015 mol

The number of moles of triprotic acid in the solution:

n(acid) = M(acid) × V(acid) = 0.1 mol/dm³ × 90 cm³ = 0.1 mol/dm³ × 0.090 dm³ = 0.009 mol

The reaction between acid and base is as follows:

3X(OH)₂ + 2H₃Y = X₃Y₂ + 6H₂O

From here:

n(base) = 0.015 mol / 3 = 0.005 mol

n(acid) = 0.009 mol / 2 = 0.0045 mol

As a result, after the neutralization, 0.005 mol - 0.0045 mol = 0.0005 mol of the base are left in the solution and 0.0005 mol × 2 = 0.001 mol of hydroxide ions are present.

The molar concentration of the hydroxide ion in the solution equals:

M(OH^-) = n (OH^-) / V(solution) = 0.001 mol / (0.150 dm³ + 0.090 dm³) = 0.001 mol / 0.24 dm³ = 0.004 M

Answer: 0.004 M