posted by Joseph .
22 mL of 0.37 mol per litre acetic acid is titrated by way of a standardized 0.29 mol/L KOH solution. Calculate the pH of the solution after roughly 18 mL of the solution of KOH is added. The Ka of acetic acid proves to be 1.8 x 10^-5
The equation is
CH3COOH + NaOH ==> CH3COONa + HOH
mol CH3COOH initially = M x L = ??
mol NaOH initially = M x L
Look to see how much has reacted, how much sodium acetate (the salt) has been formed, and how much CH3COOH remains unreacted. You will find that there is salt and an excess of CH3COOH and this constitutes a buffer solution (a weak acid and its conjugate base). Then use the Henderson-Hasselbalch equation and calculate pH. Post your work if you get stuck.
dear DR Bod
I had the same problem and worked it out as follows:
the equation: CH3COOH + NaOH ==> CH3COONa + H20
MOL CH3COOH = M x L
MOL NaON = M x L
now, to find how much sodium acetate formed, do i have to add the moles of NaOH and CH3COOH together? and how do i find the amount of CH3COOH that has not reacted?