Question

olutions that contain a weak acid, HA, and its conjugate base, A−, are called buffer solutions because they resist drastic changes in pH. When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. For buffers containing a weak acid, the principal reaction is HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq) Unbuffered, a weak acid would ionize and the net reaction would proceed forward to reach equilibrium. However, when a significant amount of conjugate base is already present, the extent of the net reaction forward is diminished. Thus, equilibrium concentrations of HA and A− are approximately equal to their initial concentrations. Visualizing buffers You can visualize a buffer solution containing approximately equal concentrations of HA and A− in action using the following equation, where Ka is the equilibrium constant: Ka[H3O+]==[H3O+][A−][HA]Ka[HA][A−] When the concentrations of HA and A− are relatively large, the addition of a small amount of acid or base changes their concentration negligibly. The result is that the ratio [HA]/[A−] maintains a value approximately equal to 1, and the [H3O+] is approximately equal to the Ka value under buffer conditions. Part A Calculate the pH of the solution made by adding 0.50 mol of HOBr and 0.30 mol of KOBr to 1.00 L of water. The value of Ka for HOBr is 2.0×10−9. Express your answer numerically

Answer 1

Answer:

pH = 8.477

Explanation:

• KOBr → K+  +  OBr-

∴ ni    0.3       -          -

  nf      -        0.3       0.3

• HOBr  ↔ H3O+  +    OBr-

∴ ni   0.50          -              0.3            

  nf  0.5 - X        X           0.3 + X

∴ Ka = 2.0 E-9 = ([H3O+]*[OBr-]) / [HOBr]

⇒ Ka = 2.0 E-9 = ((X/1L)*(0.3 + X)/1 L) / ((0.5 - X)/1L)

⇒ 2.0 E-9 = ( 0.3X + X² ) / (0.5 - X)

⇒ X² + 0.3X - 1 E-9 = 0

⇒ X = 4.333 E-9 M

according Henderson-Hauselbach:

• pH = pk + Log [A-] / [HA]

∴ [OBr-] = 0.3 mol/ 1 L + 4.333 E-9 M = 0.300 M

∴ [HOBr] = 0.5 mol / 1 L - 4.33 E-9 M = 0.500 M

∴ pKa = - log Ka = - Log ( 2.0 E-9 ) = 8.6989

⇒ pH = 8.6989 + Log ( 0.300/0.500 )

⇒ pH = 8.477