I keep getting the last part of this question wrong.
Strong base is dissolved in 765 mL of 0.200 M weak acid (Ka = 4.63 x 10^-5) to make a buffer with a pH of 4.18. Assume that the volume remains constant when the base is added. HA(aq) + OH-(aq) -. H2O(l) + A-(aq). I found that the pKa of the acid = 4.33 and the number of moles of acid initially present is 0.153 mol HA. [A-]/[HA] = 0.71 which are correct but i can't find how many moles of strong base were initially added!
(x/0.153-x) = .71 ----> 0.128 mol of OH is incorrect
and 0.153 [which i really thought would be correct] is incorrect as well!
y = 0.71(0.153 mol - y)
y = 0.153 mol
PLEASE help me!! this problem is frustrating me
You're on the right track; you're using the wrong signals.
...........HA + OH^- ==> A^- + H2O
initial..0.153..0.........0......0
added...........x..................
change.....-x...-x........x.......x
equil...0.153-x..0.........x
4.18 = 4.33 + log(x/153-x)
Solve for x. I get about 0.06 but you need to do it more accurately. That makes
acid left = 0.153-0.06 and base = 0.06.
Check it by substituting base and acid numbers into the HH equation and make sure you get 4.18.
Now that I look over your work again, I agree with what you've done except I used 0.708 instead of 0.71. I think you just made a math error because
(x/0.153-x) = 0.708 is right to that point. It's the 128 that isn't right.
To find the number of moles of the strong base initially added to the solution, we can use the information given in the problem.
Let's assume that the number of moles of the strong base initially added is y mol.
The balanced equation for the reaction between the weak acid (HA) and the strong base (OH-) is:
HA(aq) + OH-(aq) → H2O(l) + A-(aq)
Initially, the number of moles of HA present is 0.153 mol. When the strong base is added, it reacts with some of the weak acid to form water and the conjugate base A-.
The equation for the weak acid dissociation (Ka expression) is:
Ka = [A-][H+]/[HA]
Since we are given the Ka value, we can calculate the concentration of [A-]/[HA]:
Ka = [A-][H+]/[HA]
4.63 x 10^-5 = (0.71)(0.153-y)/(y)
Now, let's solve for y:
4.63 x 10^-5 = (0.10923 - 0.71y)/(y) [substituting 0.71 as [A-]/[HA] and 0.153 as the initial moles of HA]
Cross-multiplying:
(4.63 x 10^-5)y = 0.10923 - 0.71y
Re-arranging the equation:
0.71y + (4.63 x 10^-5)y = 0.10923
(0.71 + 4.63 x 10^-5)y = 0.10923
Now, solve for y:
y = 0.10923 / (0.71 + 4.63 x 10^-5)
y ≈ 0.10923 / 0.71
y ≈ 0.154 mol
Therefore, the number of moles of the strong base initially added is approximately 0.154 mol.
Please note that due to rounding errors, the calculated value may vary slightly, but it should be close to 0.154 mol.
To determine the initial moles of strong base added, we can use the equation for the buffer system (HA and A-) and their respective concentrations.
Let's denote the initial moles of strong base added as "y."
The initial moles of weak acid (HA) present in the solution is given as 0.153 mol.
The concentration of HA is 0.200 M, so the initial volume of solution can be calculated as:
Volume (V) = Moles (n) / Concentration (C) = 0.153 mol / 0.200 M = 0.765 L = 765 mL
Now, using the Henderson-Hasselbalch equation for a buffer solution:
pH = pKa + log([A-] / [HA])
Given that the pH = 4.18 and the pKa = 4.33, we can rearrange the equation to solve for ([A-] / [HA]):
[A-] / [HA] = 10^(pH - pKa) = 10^(4.18 - 4.33) = 0.71
Now, using the law of mass action expressions for the weak acid and the strong base:
HA + OH- ↔ H2O + A-
The initial concentration of OH- can be represented as "y" since we are trying to find it.
The concentration of A- can be calculated as:
[A-] = ([A-] / [HA]) * [HA] = 0.71 * 0.153 mol = 0.109 mol
In the balanced equation, we see that the moles of OH- is equal to the moles of A-. Therefore, the initial moles of OH- can be calculated as:
y = 0.109 mol
So, the initial moles of strong base added are 0.109 mol.
However, it seems like your answer differs from this calculation. It's possible that there may be a mistake or discrepancy in the given information or calculations. I would recommend double-checking the provided data and recalculating the steps to ensure accuracy. If you are still having trouble, it might be helpful to consult your instructor or a classmate for further assistance.