Question

Please answer both questions 3 and 4. Thank you!

3) The half-life for the 1st order decay of a neurotoxin is found to be 2.5 hrs. If a patient reports to have ingested a initial quantity equivalent to 200 UM of toxin into the blood, how long would one need to wait before the level dropped to an acceptable level of 1.0 UM? 4) 100ml of 0.100M Sulfuric Acid(H2SO4) is added to 200ml of 0.15M Barium Hydroxide (Ba(OH)2), what is the resulting pH?

Answer 1

Question number (3)

We know that for first Order reaction.

k = 0.693/t​​​​​​1/2 .........(1)

t = (2.303/k)log(a/a-x). .......(2)

Where,

k = rate constant

t​​​​​​1/2 = half life period = 2.5hrs

t = total time

a = initial consultation of reactant = 200microMolar

a-x = final concentration of reactant = 1micromolar

Putting the value of k in equation number (2)

t = (2.303t1/2/0.693)log(a/a-x)

Using the given values

t = (2.303×2.5hrs/0.693)log(200/1)

t = (8.31hrs)×log(200)

t = (8.31hrs)(2.3010)

t = 19.1hrs

Question (4) we know that,

number of mole of X = (concentration of x)(volume of X)

1000mL = 1L

1M = 1mol/L

Step(1) Calculation For Number of mole of sulfuric acid.

Volume of sulfuric acid = 100mL = 0.1L

Concentration of sulfuric acid = 0.100M = 0.100mol/L

Using the formula

Number of mole of Sulfuric acid = (0.100mol/L)(0.1L)

Number of mole of Sulfuric acid = 0.010mol

Step(2) Calculation For number of mole of Barium hydroxide.

Volume of Ba(OH)2 = 200mL = 0.2L

Concentration of Ba(OH)2 = 0.15M = 0.15mol/L

Using the formula

Number of mole of Ba(OH)2 = (0.15mol/L)(0.2L)

Number of mole of Ba(OH)2 = 0.03mol

Step(3) Calculation For Unreacted number of mole of Ba(OH)2 .

Chemical reaction between sulfuric acid and Barium hydroxide will be given as fofollows.

H2SO​​​​4  + Ba(OH)2 --------> Ba(SO)4  + 2H2O

From the stoichiometry of chemical reaction.

1mol sulfuric acid completely reacted with 1mol Barium hydroxide.

So

0.01mol sulfuric acid completely reacted with 0.01mol Barium hydroxide.

Unreacted Barium hydroxide = ( total Barium hydroxide) - (reacted Barium hydroxide)

Unreacted Barium hydroxide = 0.03mol - 0.01mol

Unreacted Barium hydroxide = 0.02mol

Step(4) Calculation For number of mole of OH​​​​​-

Ba(OH)2  ---------------> Ba​​​​​2+  + 2OH-

From the stoichiometry of chemical reaction.

1mole Barium hydroxide gives 2 mole of OH​​​​​- .

So

0.02mole Barium hydroxide given 0.04mol of OH​​​-

Number of mole of OH​​​​​- = 0.04mol

Step(5) Calculation For concentration of OH​​​​​-

Concentration = (number of mole )/(total volume)

Total volume = volume of sulfuric acid + volume of Barium hydroxide

Total volume = 100mL + 200mL = 300mL = 0.3L

So,

Concentration of OH​​​​​- = [OH​​​​​- ] = (0.04mol)/(0.3L)

[OH​​​​​-​] = 0.012mol/L = 0.012M

Step(6) Calculation For P​​​​​​H .

We know that,

P​​​​​​H = -log[H+ ] .......(1)

Or

P​​​​​​H = 14 + log[OH​​​​​- ] .......(2)

Using the equation number (2)

P​​​​​​H = 14 + log(0.012)

P​​​​​​H = 14 - 1.92

P​​​​​​H = 12.08