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

pH = 8.477

Explanation:

∴ ni    0.3       -          -

  nf      -        0.3       0.3

∴ 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:

∴ [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

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years