A buffer is prepared by adding 300.0 mL of M NaOH to 400.0 ML of 2.0 M CH3COOH. What is the pH of this buffer? ks=1.8X10^-5
You didn't type in the M of the NaOH.
To find the pH of the buffer, we need to calculate the concentration of the acid (CH3COOH) and its conjugate base (CH3COO-) after mixing.
First, let's calculate the moles of NaOH added. We have a volume of 300.0 mL and the concentration is given as M (moles per liter). So, we can convert the volume into liters by dividing by 1000:
300.0 mL = 300.0/1000 = 0.3 L
To calculate the moles of NaOH, we'll multiply the volume (in liters) by the concentration:
moles of NaOH = 0.3 L x M NaOH
Next, we need to calculate the moles of CH3COOH present initially. We have a volume of 400.0 mL and a concentration of 2.0 M:
moles of CH3COOH = 400.0 mL x 2.0 M / 1000 mL
Now, let's calculate the moles of CH3COO- formed when NaOH reacts with CH3COOH. Since NaOH is a strong base and CH3COOH is a weak acid, the reaction is a complete neutralization:
moles of CH3COO- = moles of NaOH
Now, let's calculate the total moles of CH3COOH and CH3COO- in the buffer solution. We'll add the initial moles of CH3COOH and the moles of CH3COO- formed:
total moles of CH3COOH = moles of CH3COOH initial - moles of CH3COO-
total moles of CH3COO- = moles of CH3COO-
Now, we can calculate the concentration of CH3COOH and CH3COO- in the buffer solution. Since the total volume of the solution is the sum of the volumes of CH3COOH and NaOH, which is 400.0 mL + 300.0 mL = 700.0 mL = 0.7 L:
concentration of CH3COOH = total moles of CH3COOH / 0.7 L
concentration of CH3COO- = total moles of CH3COO- / 0.7 L
Finally, we can calculate the pH of the buffer using the Henderson-Hasselbalch equation:
pH = pKa + log([CH3COO-] / [CH3COOH])
Since the Ks value for CH3COOH (acetic acid) is given, we can use its negative logarithm (-log10):
pKa = -log10(Ks)
Substituting the values into the Henderson-Hasselbalch equation will give us the pH of the buffer.