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
I got the pH is 4.79 but i can't figure out the grams of KOH
the problem says how many grams of KOH must be added to .500 L of the buffer to change the pH by .125 units?
I got .221 g KOH but it says that's wrong
I worked this yesterday. Let me see if I can find it.
To calculate the grams of KOH needed to change the pH of the buffer, you will need to follow a series of steps. Let's go through the process together:
Step 1: Find the initial concentrations of acetic acid (CH3COOH) and its conjugate base sodium acetate (CH3COONa) in the buffer solution.
- The concentration of acetic acid (CH3COOH) will be equal to the concentration of sodium acetate (CH3COONa).
- Given that the volume of sodium acetate is 0.500 L and the concentration is 0.400 M, the number of moles of sodium acetate (CH3COONa) can be calculated as follows:
Moles of CH3COONa = concentration × volume = 0.400 M × 0.500 L = 0.200 moles
- Since acetic acid and sodium acetate have a 1:1 molar ratio, the concentration of acetic acid will also be 0.400 M.
Step 2: Calculate the initial concentrations of H+ and CH3COO- in the buffer solution.
- According to the Henderson-Hasselbalch equation, the pH of the buffer can be calculated using the equation:
pH = pKa + log([CH3COO-] / [CH3COOH])
- Given that the pKa of acetic acid is the negative logarithm of the Ka value, pKa = -log(Ka) = -log(1.80 × 10^-5) = 4.74.
- Also, from the problem statement, we know that the pH is initially 4.79. We can use the Henderson-Hasselbalch equation to calculate the ratio of [CH3COO-] / [CH3COOH]:
4.79 = 4.74 + log([CH3COO-] / [CH3COOH])
- Rearranging the equation gives:
log([CH3COO-] / [CH3COOH]) = 4.79 - 4.74 = 0.05
- Taking the antilog of both sides:
[CH3COO-] / [CH3COOH] = 10^0.05 = 1.122
- This means that for every molecule of CH3COOH, there are approximately 1.122 molecules of CH3COO-. This allows us to calculate their concentrations:
[CH3COO-] = 1.122 × [CH3COOH] = 1.122 × 0.400 M = 0.449 M
[CH3COOH] = 0.400 M
- Therefore, the initial concentration of H+ is equal to 0.400 M, and the initial concentration of CH3COO- is 0.449 M.
Step 3: Determine the concentration of CH3COO- after changing the pH by 0.125 units.
- Since the pH changes by 0.125 units, the new pH will be 4.79 + 0.125 = 4.915.
- We will use the new pH value, pKa value, and the Henderson-Hasselbalch equation to find the new ratio of [CH3COO-] / [CH3COOH]:
4.915 = 4.74 + log([CH3COO-] / [CH3COOH])
- Rearranging the equation:
log([CH3COO-] / [CH3COOH]) = 4.915 - 4.74 = 0.175
- Taking the antilog of both sides:
[CH3COO-] / [CH3COOH] = 10^0.175 = 1.459
- This means that for every molecule of CH3COOH, there are approximately 1.459 molecules of CH3COO-. This allows us to calculate their concentrations:
[CH3COO-] = 1.459 × [CH3COOH] = 1.459 × 0.400 M = 0.584 M
[CH3COOH] = 0.400 M
Step 4: Calculate the moles of CH3COOH and KOH needed.
- Since the volume of the buffer solution is given as 0.500 L, we can calculate the moles of CH3COOH and KOH needed using their concentrations:
Moles of CH3COOH = concentration × volume = 0.400 M × 0.500 L = 0.200 moles
Moles of KOH = concentration × volume = x M × 0.500 L = 0.500x moles
Step 5: Set up an equation to find the moles of KOH needed.
- The moles of CH3COOH react with the moles of KOH in a 1:1 ratio:
0.200 moles CH3COOH = 0.500x moles KOH
Step 6: Solve for x to find the concentration of KOH.
- Rearrange the equation:
0.500x = 0.200
- Solve for x:
x = 0.200 / 0.500 = 0.400 M
Step 7: Calculate the grams of KOH needed.
- Using the concentration of KOH, we can calculate the molar mass of KOH to find the grams needed:
Molar mass of KOH = K (39.10 g/mol) + O (16.00 g/mol) + H (1.01 g/mol) = 56.11 g/mol
- Calculate the grams of KOH needed using the mole-to-grams conversion:
Grams of KOH = moles × molar mass = 0.400 M × 0.500 L × 56.11 g/mol = 11.22 g
Therefore, the correct amount of KOH needed to change the pH by 0.125 units in a 0.500 L buffer is 11.22 grams.