An aqueous solution of sodium hydroxide is standardized by titration with a 0.104 M solution of hydrobromic acid. If 13.5 mL of base are required to neutralize 20.2 mL of the acid, what is the molarity of the sodium hydroxide solution?

0.156 M

Explanation:

The neutralization reaction between hydrobromic acid (HBr) and sodium hydroxide (NaOH) is the following:

HBr(ac) + NaOH(aq) → NaBr(aq) + H₂O(l)

As we can see, 1 mol of HBr reacts with 1 mol NaOH. During a titration, at the equivalence point, the number of moles of NaOH added reacts completely with the number of moles of HBr:

moles of NaOH = moles of HBr

The moles of each reactant is calculated as the product of the molarity (M) and the volume (V):

M(NaOH) x V(NaOH) = M(HBr) x V(HBr)

We have the following data:

base: V(NaOH) = 13.5 mL

acid: M(HBr) = 0.104 M;  V(HBr) = 20.2 mL

Thus, we calculate M(NaOH) from the equivalent point equation:

M(NaOH) = (M(HBr) x V(HBr))/(V(NaOH) = (0.104 M x 20.2 mL)/(13.5 mL) = 0.1556 M ≅ 0.156 M

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

am i too late?

explanation: