PLEASE HELP ASAP

Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH

uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

7 0

3 years ago

of an unknown protein are dissolved in enough solvent to make 5.00mL of solution. The osmotic pressure of this solution is measu

krok68 [10]

The question is incomplete . The complete question is :

100 mg of an unknown protein are dissolved in enough solvent to make 5.00mL of solution. The osmotic pressure of this solution is measured to be 0.107atm at 25.0°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits.

Answer: The molar mass of the protein is $4.57\times 10^3g/mol$

Explanation:

$\pi =CRT$

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 0.107 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (protein) = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 5.00 mL

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$0.107=1\times \frac{0.1\times 1000}{\text{Molar mass of insulin}\times 5.00}\times 0.0821\text{ Latm }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of protein}$