Answer to Question #196256 in Chemistry for Kenan

Question #196256

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!


1
Expert's answer
2021-05-24T06:30:41-0400

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

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