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hope it's help!
Answer:
1. 6.005 g
2. 22.9 mL
3. Until the mixtures becomes homogeneous.
Explanation:
A buffer is a solution where a weak acid is in equilibrium with its conjugate base (its anion) or a weak base is in equilibrium with its conjugate base (its cation). The buffer remains the pH almost unaltered because it shifts the equilibrium if an acid or base is added.
1. The pH of a buffer can be calculated by the Henderson-Hasselbalch equation:
pH = pKa + log[A⁻]/[HA]
Where [A⁻] is the concentration of the conjugate base (the anion) of the acid, and HA is the acid concentration.
5.10 = 4.76 + log[A⁻]/[HA]
log[A⁻]/[HA] = 5.10 - 4.76
log[A⁻]/[HA] = 0.34
[A⁻]/[HA] =
[A⁻]/[HA] = 2.1878
Because the volume is the same, we can replace the concentration by the number of moles (n):
nA⁻/nHA = 2.1878
nA⁻ = 2.1878*nHA
The total number of moles of the substances in the buffer is: 0.200 mol/L * 0.5 L = 0.1 mol
nA⁻ + nHA = 0.1
2.1878*nHA + n HA = 0.1
3.1878nHA = 0.1
nHA = 0.0314 mol
nA⁻ = 0.0686 mol
The total number of moles of acetic acid needed is 0.1 mol (both substances may be from it):
m = MW*mol
m = 60.05*0.1 = 6.005 g
2. NaOH must react with acetic acid to form the anion, so for a 1:1 reaction, it will be needed 0.0686 mol of NaOH:
V = mol/concentration
V = 0.0686/3
V = 0.0229 L = 22.9 mL
3. The buffer must be a homogeneous solution, it means that it can't be noticed phases in the buffer, so the flask must be inverted until all the buffer is diluted in water, and it will be noticed when the solution becomes homogenous.
Density = mass / volume
0.601 = m / 7.25
m = 0.601 x 7.25
m =4.35725 g
hope this helps!
Answer:
4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL
Explanation:
A 3%(w/v) solution contains 3g of solute (In this case, Na2CO3) in 100mL of solution.
Assuming we require 100mL of solution we must add 3g of Na2CO3. The reactant that is available is its dihydrate, with molar mass:
106g/mol + 2*MW H2O
106g/mol + 2*18g/mol = 142g/mol
That means the mass of Na2CO3.2H2O that must be added to prepare the solution is:
3g Na2CO3 * (142g/mol Na2CO3.2H2O / 106g/mol Na2CO3) =
<h3>4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL</h3>