A buffer is prepared by mixing 205 mL of .452 M HCl and .500 L of .400 M sodium acetate. (Ka =1.80 x 10^-5)
a) What is the pH?
b) How many grams of KOH must be added to .500 L of the buffer to change the pH by .125 units?
a is solved using the Hendrson-Hasselbalch equation.
b. Add 0.125 to pH from a. I think that will be about 4.925 but you need to confirm that. Then find mmoles acetic acid and mmoles acetate that you have at the beginning of problem b, divide by mL to find M of each, and multiply by 500 mL to find mmoles of each in 0.5 L.
Adding KOH will mean (acetate) = mmoles+x and (acetic acid) will be mmoles-x. Plug that into the Henderson-Hasselbalch equation and solve for x. Then convert that to grams KOH for the 0.5 L buffer solution. Post your work if you get stuck.
I set up the Hendrson-hasselbalch equation and i got 4.95=log((.4-x)/(1.1+x) but now i'm stuck on how to figure out x
To answer these questions, we need to consider the reaction between HCl and sodium acetate.
First, let's calculate the number of moles of HCl and sodium acetate in the solution.
a) What is the pH?
1. Calculate the moles of HCl:
Moles of HCl = Molarity × Volume
Moles of HCl = (0.452 M) × (0.205 L) = 0.09286 moles
2. Calculate the moles of sodium acetate:
Moles of sodium acetate = Molarity × Volume
Moles of sodium acetate = (0.400 M) × (0.500 L) = 0.200 moles
3. Use the moles of the weak acid (sodium acetate) and the weak base (acetate ion) to determine the equilibrium concentrations:
Calculate the initial concentration of acetate ion (CH3COO-) in the buffer solution:
Initial concentration of sodium acetate = Moles of sodium acetate / Total volume of the buffer
Initial concentration of sodium acetate = 0.200 moles / (0.205 L + 0.500 L) = 0.235 M
4. Write the equilibrium equation of the sodium acetate buffer:
CH3COOH + H2O ⇌ CH3COO- + H3O+
In this reaction, sodium acetate acts as the weak base that reacts with water to form acetate ions (CH3COO-) and hydronium ions (H3O+).
5. Use the equilibrium expression to find the concentration of hydronium ions ([H3O+]):
Ka = [CH3COO-] × [H3O+] / [CH3COOH]
Since the initial concentration of [CH3COO-] is the same as [H3O+] and [CH3COOH] will be negligible compared to [CH3COO-], we can simplify the equation to:
Ka = [H3O+]
Therefore, [H3O+] = Ka = 1.80 × 10^-5
6. Calculate the pH using the concentration of hydronium ions ([H3O+]):
pH = -log [H3O+]
pH = -log (1.8 × 10^-5)
pH ≈ 4.744
b) How many grams of KOH must be added to .500 L of the buffer to change the pH by .125 units?
To calculate the number of grams of KOH required, we need to consider the change in moles of the buffer components.
1. Calculate the change in moles of hydronium ions ([H3O+]):
Δ[H3O+] = antilog(pH change) = antilog(0.125)
Δ[H3O+] ≈ 1.35
2. Determine the change in moles of acetate ions:
Since [H3O+] and [CH3COO-] are initially the same, the change in moles of acetate ions will also be Δ[H3O+].
3. Calculate the moles of KOH required to react with the change in moles of hydronium ions:
Moles of KOH = Δ[H3O+] = 1.35 moles
4. Convert moles of KOH to grams using the molar mass of KOH:
Moles of KOH × Molar mass of KOH = Grams of KOH
Please note that this answer assumes ideal behavior and complete reaction of the components. In reality, there may be slight deviations due to factors such as incomplete reaction or non-ideal behavior.