Need help with question. Calculate the pH of a buffer prepared by adding 100mL of 0.0500M NaOH to 100mL of 0.175M CH3COOH.
so far I have calculated mols of
CH3COOH---->H+ + CH3OO-
CH3COOH=0.1L*0.175mol/1L=0.0175mol
OH-=0.1L*0.0500/1L=0.00500mol
Not sure were to go from here, or if the above is right
Millimoles is a little easier to work with in these problems; it keeps all those leading zeros out of the way.
100 mL x 0.175M CH3COOH = 17.5 mmoles.
100 mL x 0.05M NaOH = 5 mmoles.
.......NaOH + CH3COOH ==> CH3COONa + H2O
init...5.0....17.5.........0..........0
change.-5.0...-5.0.......+5.0
equil...0....12.5.........5.0
So you have CH3COOH and CH3COONa (a weak acid and its salt which gives you a buffer). Use the Henderson-Hasselbalch equation to solve for pH. I get about 4.3 or so but you need to do it more accurately.
pH = pKa + log[(acetate)/(acid)]
How do i solve if i am not aloud to use henderson-hasselbach?
You're on the right track! Calculating the number of moles of acetic acid (CH3COOH) and hydroxide ions (OH-) is the first step.
Next, you can use the concept of a buffer solution, which is a solution that resists changes in pH when small amounts of acid or base are added. In this case, you have a mixture of acetic acid (a weak acid) and sodium hydroxide (a strong base). The reaction between the weak acid and strong base will result in the formation of water and the acetate ion (CH3COO-):
CH3COOH + NaOH → H2O + CH3COO-Na+
Since acetic acid is a weak acid, it does not fully ionize in water. Therefore, we need to consider the dissociation of acetic acid in water using the equilibrium constant expression:
CH3COOH ⇌ H+ + CH3COO-
The equilibrium constant for this dissociation reaction is called the acid dissociation constant (Ka). The expression for Ka is:
Ka = [H+][CH3COO-] / [CH3COOH]
We can rearrange this expression to solve for [H+]:
[H+] = Ka * [CH3COOH] / [CH3COO-]
In this case, the concentration of CH3COOH is given as 0.0175 mol, but the concentration of CH3COO- is not provided. However, we can calculate it using the number of moles of NaOH that reacted with acetic acid. Since sodium hydroxide is a strong base, it completely dissociates in water into sodium ions (Na+) and hydroxide ions (OH-):
NaOH → Na+ + OH-
We have determined the number of moles of NaOH as 0.00500 mol. Since NaOH and CH3COOH react in a 1:1 ratio, the number of moles of CH3COO- formed will also be 0.00500 mol.
Now we have all the values needed to calculate the pH:
[H+] = Ka * [CH3COOH] / [CH3COO-]
[H+] = Ka * 0.0175 mol / 0.00500 mol
To find Ka, you can look it up in a table of acid dissociation constants. For acetic acid, Ka is approximately 1.8 x 10^-5.
[H+] = (1.8 x 10^-5) * 0.0175 mol / 0.00500 mol
Calculating this expression will give you the concentration of H+ ions. Finally, you can convert the concentration of H+ ions to pH using the equation:
pH = -log[H+]
Plug in the value you calculated for [H+] and solve for pH to find the pH of the buffer solution.