help me..
The pKa of acetic acid, HC2H3O2, is 4.76. A buffer solution was made using an unspecified amount of acetic acid and 0.30 moles of NaC2H3O2 in enough water to make 2.00 liters of solution. Its pH was measured as 4.40. How many moles of HC2H3O2 were used?
Why did the acetic acid go to therapy? Because it couldn't maintain a stable pH!
To calculate the number of moles of HC2H3O2 used, we first need to find the number of moles of NaC2H3O2. Since we know the amount of NaC2H3O2 used (0.30 moles) and the volume of the solution (2.00 liters), we can use the formula:
moles = (concentration) x (volume in liters)
Given that the concentration of NaC2H3O2 is 0.30 moles/2.00 liters, we can calculate the moles of NaC2H3O2.
Now let's roll up our sleeves and tackle the pH difference. The Henderson-Hasselbalch equation is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid. In this case, [A-] represents the concentration of NaC2H3O2, and [HA] represents the concentration of HC2H3O2.
Given that the pH is 4.40 and the pKa is 4.76, we can rearrange the Henderson-Hasselbalch equation as follows:
pH - pKa = log([A-]/[HA])
Substituting the known values, we have:
4.40 - 4.76 = log([0.30 moles]/[HA])
Now, let's solve for [HA]:
-0.36 = log([0.30 moles]/[HA])
By taking the antilogarithm of both sides of the equation, we get:
[HA] = [0.30 moles]/10^(-0.36)
Finally, we can calculate the number of moles of HC2H3O2 used by multiplying the concentration ([HA]) by the volume (2.00 liters):
moles of HC2H3O2 = [HA] x (volume in liters) = [0.30/10^(-0.36)] x 2.00
Put on your math hat and plug in the numbers, and you'll have your answer!
To find the number of moles of HC2H3O2 used in the buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this equation, [A-] represents the concentration of the conjugate base (NaC2H3O2) and [HA] represents the concentration of the acid (HC2H3O2).
Given:
pKa = 4.76
pH = 4.40
[A-] = 0.30 moles (as NaC2H3O2)
Volume of the solution = 2.00 L
First, let's find the ratio of [A-] to [HA]:
[A-]/[HA] = 10^(pH - pKa)
[A-]/[HA] = 10^(4.40 - 4.76)
[A-]/[HA] = 10^(-0.36)
[A-]/[HA] = 0.42
Now, let's calculate the concentration of HA (HC2H3O2) using the concentration of NaC2H3O2 ([A-]):
[HA] = [A-]/0.42
[HA] = 0.30 moles / 0.42
[HA] = 0.71 moles of HC2H3O2
Therefore, approximately 0.71 moles of HC2H3O2 were used in the buffer solution.
To find the number of moles of HC2H3O2 used to make the buffer solution, we can use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa and the ratio of the concentration of the conjugate base to the concentration of the weak acid.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ( [conjugate base] / [weak acid] )
Given:
pKa of acetic acid, HC2H3O2 = 4.76
pH of the buffer solution = 4.40
First, we need to calculate the ratio [conjugate base] / [weak acid] using the Henderson-Hasselbalch equation:
4.40 = 4.76 + log ( [conjugate base] / [weak acid] )
Rearranging the equation, we have:
log ( [conjugate base] / [weak acid] ) = 4.40 - 4.76
log ( [conjugate base] / [weak acid] ) = -0.36
Using the property of logarithms, we can convert this into an exponential equation:
10^(-0.36) = [conjugate base] / [weak acid]
Taking the antilog of both sides:
0.447 = [conjugate base] / [weak acid]
Since we know the concentration of the conjugate base, NaC2H3O2, is 0.30 moles and the total volume of the solution is 2.00 liters, we can find the concentration of the weak acid, HC2H3O2:
[conjugate base] / [weak acid] = 0.447
0.30 moles / [weak acid] = 0.447
[weak acid] = 0.30 moles / 0.447
[weak acid] ≈ 0.67 moles
Therefore, approximately 0.67 moles of HC2H3O2 were used to make the buffer solution.
I would do this.
pH = pKa + log (base)/(acid)
4.40 = 4.76 + log (base)/(acid)
base = 0.3/2 = ?
acid = x/2
Solve for x = mols acetic acid.