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You added 20 ml of 0.20M solution of Ba(OH)2 to a 0.10M solution of HCl(aq).

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Final answer:

When a solution of Ba(OH)2 is mixed with HCl, a neutralization reaction occurs, forming water and BaCl2. To determine if a precipitate of barium sulfate will form, we need to calculate the ion product (Q) and compare it to the solubility product (Ksp) of BaSO4. In this case, Q is greater than Ksp, indicating that a precipitate will form.

Step-by-step explanation:

When you mix a solution of Ba(OH)2, which is a base, with a solution of HCl, which is an acid, a neutralization reaction occurs. In this reaction, Ba(OH)2 and HCl react to form water and a salt, which is BaCl2. The balanced chemical equation for this reaction is: Ba(OH)2 + 2HCl → BaCl2 + 2H2O.

To determine if a precipitate of barium sulfate (BaSO4) will form, we need to compare the solubility of barium sulfate with the concentrations of barium ions (Ba2+) and sulfate ions (SO42-) in the solution. If the ion product (Q) of barium sulfate is greater than the solubility product (Ksp), a precipitate will form.

From the given information, we know that 20 mL of a 0.20 M solution of Ba(OH)2 is added to a 0.10 M solution of HCl. To calculate the concentrations of Ba2+ and SO42-, we need to use the dilution equation. Given that the volume of the final solution is 20 mL + 100 mL = 120 mL, we can calculate the new molarities:

  1. Ba2+: (0.20 M * 20 mL) / 120 mL = 0.033333 M
  2. SO42-: 0.10 M

With the concentrations of Ba2+ and SO42- in the solution, we can now calculate the ion product (Q) of barium sulfate using the equation Q = [Ba2+][SO42-]. Substituting the molarities of Ba2+ and SO42- into the equation:

Q = (0.033333 M)(0.10 M) = 0.003333

The solubility product (Ksp) of barium sulfate is approximately 1.1 x 10^-10. Since the ion product (Q) is greater than the solubility product (Ksp), a precipitate of barium sulfate will form.

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