ezto.mheducation.com/hm.tpx h. Ex. 154ab- Weak Acid Equilibria 2 out of 3 attemp

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ezto.mheducation.com/hm.tpx h. Ex. 154ab- Weak Acid Equilibria 2 out of 3 attempts The beakers shown contain 0.300 L of aqueous solutions of a moderately weak acid HY. Each particle represents 0.0100 mol; solvent molecules are omitted for clarity. Protons are smaller and light blue. Oxygen atoms are red. Therefore, hydronium ions are red with three light blue hydrogen atoms. 900 0 D 0 (a) The reaction in beaker A is at equilibrium. Calculate Q for B, C, and D to determine which, if any, are also at equilibrium. 2 B ec.. [ ] 2D-?????? ? References eBook & Resources Multipart Answer

Explanation / Answer

weak acid = HA

HA + H2O <----> H3O+(aq) + A-(aq)

A) at equilibrium,

   [H3O+] = [A-] = n/v = 4*0.01/0.3 = 0.133 M

   [HA] = n/v = 12*0.01/0.3 = 0.4 M

Ka = [H3O+][A-]/[HA]

    = (0.133*0.133)/0.4

    = 0.044

B) [H3O+] = [A-] = n/v = 2*0.01/0.3 = 0.0667 M

   [HA] = n/v = 8*0.01/0.3 = 0.267 M

QB = [H3O+][A-]/[HA]

    = (0.0667*0.0667)/0.267

    = 0.0166

C)

[H3O+] = [A-] = n/v = 2*0.01/0.3 = 0.0667 M

   [HA] = n/v = 6*0.01/0.3 = 0.2 M

QC = [H3O+][A-]/[HA]

    = (0.0667*0.0667)/0.2

    = 0.022

d)

[H3O+] = [A-] = n/v = 2*0.01/0.3 = 0.0667 M

[HA] = n/v = 4*0.01/0.3 = 0.133 M

QC = [H3O+][A-]/[HA]

    = (0.0667*0.0667)/0.133

    = 0.0334

if Ka = Q , system is at equilibrium.

from the above data,no one at equilibrium.

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