how much 0.1F HC2H3O2 is required to make 500 ml of solution with a Ph of 4.0?
HC2H3O2 = HAc
Let Y = (HAc) needed to make pH = 4.0 or (H^+) of 1E-4 M.
.......HAc ==> H^+ + Ac^-
I......Y.......0.....0
C.....-x.......x.....x
E.....Y-x......x.....x
Ka = (H^+)(Ac^-)/(HAc)
Ka = (x)(x)/(Y-x)
You know Ka (or you can look it up), you know x, solve for Y. I will use Y as the answer but you substitute that into the following.
mL1 x M1 = mL2 x M2
500 x Y = mL2 x 0.1
Solve for mL2. That will be the volume in mL of the 0.1M HAc needed.
To calculate the amount of 0.1F HC2H3O2 required to make a 500 ml solution with a pH of 4.0, we need to consider the acid dissociation constant (Ka) of HC2H3O2 and the pH equation for weak acids.
Firstly, let's understand that HC2H3O2 is acetic acid, and its acid dissociation constant (Ka) is approximately 1.8 x 10^-5.
We can use the pH equation for weak acids, which is:
pH = pKa + log([A-]/[HA])
In this case, [A-] represents the concentration of the acetate ion (C2H3O2-) and [HA] represents the concentration of the acetic acid (HC2H3O2).
The pH is given as 4.0, so the pKa can be calculated as:
pKa = -log10(Ka) = -log10(1.8 x 10^-5) = 4.74
Next, we need to calculate the ratio of [A-]/[HA]. Since the pH is slightly above the pKa value, the ratio is close to 1:1.
Now, let's assume that 'x' moles of HC2H3O2 are needed to make 500 ml of solution. This means 'x' moles of C2H3O2- will also be present.
Since the ratio is approximately 1:1, the concentration of HC2H3O2 and C2H3O2- will be equal.
Therefore, the moles of HC2H3O2 and C2H3O2- can be expressed as x moles each.
Now, we need to convert the moles to concentration (Molarity). The volume of the solution is 500 ml or 0.5 L.
Concentration (M) = moles / volume (L)
For HC2H3O2: 0.1 M = x moles / 0.5 L
Therefore, x moles = 0.1 M × 0.5 L = 0.05 moles
So, 0.05 moles of HC2H3O2 are required to make a 500 ml solution with a pH of 4.0.
To determine how much 0.1 F HC2H3O2 is required to make a 500 ml solution with a pH of 4.0, we need to consider acid-base equilibrium and the dissociation of HC2H3O2 in water.
First, let's understand the acid dissociation reaction of HC2H3O2 in water:
HC2H3O2 ⇌ H+ + C2H3O2-
The dissociation constant of HC2H3O2, known as Ka, is used to calculate the concentration of H+ ions in solution. In this case, we know the desired pH is 4.0, which means the concentration of H+ ions should be 10^(-4.0) M.
Since HC2H3O2 is a weak acid, the concentration of H+ ions can be approximated using the equilibrium expression:
[H+] = √(Ka * [HC2H3O2])
Given that the 0.1 F HC2H3O2 has a molar concentration of 0.1 M, we can substitute the values and solve for the required amount of HC2H3O2.
√(Ka * 0.1) = 10^(-4.0)
Now, we need the Ka value for HC2H3O2, which is 1.8 × 10^(-5) M. Substituting this value into the equation, we can solve for the concentration of HC2H3O2.
√(1.8 × 10^(-5) * 0.1) = 10^(-4.0)
√(1.8 × 10^(-6)) = 10^(-4.0)
1.34 × 10^(-3.0) = 10^(-4.0)
1.34 × 10^(-3.0) = 0.00134
Therefore, to make a 500 ml solution with a pH of 4.0, you would need 0.00134 moles of 0.1 F HC2H3O2.