Calculate the pH of the resulting solution if 32.0 mL of 0.320 M HCl(aq) is added to (a) 42.0 mL of 0.320 M NaOH(aq). (b) 22.0 mL of 0.420 M NaOH(aq).

a) pH = 11.5

b) pH= 3

Explanation:

a) lets calculate the number of moles of each reactant

moles of HCl = 32/1000 * 0.32 = 0.01024 mole

moles of NaOH = 42/1000 * 0.32 = 0.01344 moles

1 moles of HCl reacts with 1 moles of NaOH , so 0.01024 mole of HCl should react with 0.01024 moles of NaOH , but there is some excess NaOH.

excess NaOH= 0.01344 -0.01024 = 0.0032 moles

[H+]=

pH= -log [3.125*10^-12) = 11.5

b)  moles of NaOH = 0.00924

excess HCl present = 0.01024 - 0.00924 =.001

so excess [H+] = 0.001

pH= -log( 0.001) = 3

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