For some reason, Gabby has to decided to collect the sulfuric acid (H2S) that ha

Question

For some reason, Gabby has to decided to collect the sulfuric acid (H2S) that has accumulated in her dead car battery. In order to repurpose it, she needs to know its concentration, so she decides to titrate a small amount of it with 0.1 M sodium hydroxide (NaOH). a. If it takes 57 mL of the 0.1 M NaOH to neutralize 10 mL of the H2S, what is the concentration of the acid? b. Woah, that is way too concentrated. How would she dilute it to only be 0.2 M H2S? Explain the procedure and show your math. A buffer is made up of acetic acid (CH3COOH) and sodium hydroxide (NaOH). a. Draw what would happen if a base (-OH) was added to the solution (the chemical equation). b. Draw what would happen if acid (H+) was added to the solution (the chemical equation). c. What is the purpose of a buffer? Patricia loves to cook, and has a lot of spices in her cabinet to prove it. Some of these spices are old though, and the labels have worn off. a. Assuming they are water soluble, how would she be able to determine the molar mass of these unknown compounds? b. The freezing point of water is 0 degree C. After Patricia adds in 0.2 grams of one of her unknown spices to 50 mL of water, the freeing point changed to -2 degree C. Assuming no ions present, what is the molar mass of the unknown spice? (for water is 1.86 C/mol Write the rate law expression for the reaction A + B rightarrow C, based on the following data: a. If the Rate units are M/sec, what is unit for k? Oh Gabby spilled an oily, non-polar substance on her hand. It's starting to burn Should she wash it off with a polar or non polar substance? Why? 6) Give an example of the following intermolecular forces (draw the interaction): a. Dispersion forces b. Dipole-dipole forces c. Hydrogen bonding d. Ion-dipole forces e. Ion-ion forces

Explanation / Answer

1)a)Let the molarity of H2SO4 be M

2NaoH + H2SO4 ----------->Na2SO4 + 2H2O

thus V1M1/n1 = V2M2/n2

0.1 x57mL/2 = 10xM

or molarity of acid = 0.285M

b) dilution to 0.2M

V1M1 = V2M2

0.285x10mL = 0.2 xV

V= 14.25 ml

Taking 10mL of 0.285 M solution and diluting it to 14.25 mL she can get 0.2 M acid

2) the buffer is made of acetic acid and sodium acetate .

CH3COOH <----------> CH3COO- + H+

CH#COONa ----------> CH3COO- + Na+

Due to common ion the dissociation of weak acetic acid is further depressed.

So [CH3COOH] in solution = [acid] and

[CH3COO-] in solution = [salt]

now the bufer has undissociated weak acid and its conjugate base

a) some OH- added

it reacts with undissociated acid as

CH3COOH + OH- ------------> CH3COO- + H2O

so no change in [H+] and thus in pH

b) when H+ is added

CH3COO_ + H+ -------> CH3COOH

again no change in [H=] in solution so pH remains the same.

3)

a)the molar mass can be calculated by measuring any colligative property like elevation in boiling point /depression in freezing point or osmotic pressure or lowering of vapor pressure.

b) Given delta Tf = 2

mass of solute = o.2 g

mass of solvent = 50g

Kf = 1.86

We have the relation as

delta Tf = Kf x m

= Kf (mass/molar mass ) x(1000/mass of solvent in g)

2 = 1.86 x(0.2/M) (1000/50)

M = 3.72 g

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