We have two liters of a buffer solution that
is 0.10 M in CH3COOH and 0.10 M in
NaCH3COO. How many moles of solid NaOH
are to be added to this solution to increase its
pH by 5.0 percent? Assume no volume change
due to the addition of solid NaOH.
I have the starting pH and what the final should be.
Initial would be 4.744 and final would be 4.98
You have the equation (CH3COOH = HAc; CH3COO^- = Ac^-).
HAc + NaOH ==> NaAc + H2O
You are starting with 0.1 M x 2 L = 0.2 mole HAc and 0.2 mole Ac^-.
We want 4.982
4.982 = 4.744 + log [(Ac^-)/(HAc)]
You must recognize that when NaOH is added, the HAc is decreased by x so the final moles = 0.2-x. At the same time, the amount of NaOH added (which is x) increases the Ac^- by x and it becomes 0.2+x. Substitute and solve for x.
4.982 = 4.744 + log [(0.2+x)/(0.2-x)]
Solve for x.
I don't know that carrying the pH to three places is justified so you need to check that out.
To solve this problem, you need to determine the change in pH and use it to calculate the amount of NaOH needed. Here's how you can approach it step by step:
Step 1: Calculate the hydrogen ion concentration (H+) for the initial and final pH values.
To convert pH to H+ concentration, you can use the equation:
[H+] = 10^(-pH)
For the initial pH of 4.744:
[H+] = 10^(-4.744)
For the final pH of 4.98:
[H+] = 10^(-4.98)
Step 2: Calculate the percent change in H+ concentration.
To find the percent change, use the formula:
Percent Change = [(Final Value - Initial Value) / Initial Value] * 100
Percent Change = [(Final [H+] - Initial [H+]) / Initial [H+]] * 100
Percent Change = [(10^(-4.98) - 10^(-4.744)) / 10^(-4.744)] * 100
Percent Change = [(0.000794 - 0.000525) / 0.000525] * 100
Step 3: Calculate the moles of CH3COOH initially present in the buffer solution.
The concentration of CH3COOH is given as 0.10 M, and you have 2 liters of the buffer solution. Therefore, the moles of CH3COOH can be calculated as:
Moles of CH3COOH = Concentration (M) * Volume (L)
Moles of CH3COOH = 0.10 M * 2 L
Step 4: Calculate the moles of NaOH needed to increase the pH by the given percent change.
To find the moles of NaOH needed, you need to consider the stoichiometric ratio between NaOH and CH3COOH, which is 1:1. This means that for every mole of NaOH added, one mole of CH3COOH will be neutralized.
Moles of NaOH needed = Moles of CH3COOH * Percent Change / 100
Moles of NaOH needed = (0.10 M * 2 L) * (Percent Change / 100)
Finally, you can substitute the value of the percent change calculated in Step 2 to compute the moles of NaOH needed to increase the pH by 5.0 percent.