pH and Buffer of Acid/Base Calculations

Hi Sektor404
Q1:
The Henderson-Hasselbach equation is
pH = pKa + log ([base]/[acid])
and the pKa of acidic acid is , i believe, 4.8

Plug the numbers into the equation we get
5.5 = 4.8 +log ([base]/[acid])

subtract 4.8 from both side we get

.7 = log ([base]/[acid])

to solve for the ratio of [base]/[acid] we raise 10 to each side of the equation. By doing so, we can get rid of the "log"

10^0.7 = 10^(log ([base]/acid]))
5.012 = [base]/[acid]

Q2:
let Vb = volume of acetate  and Va= volume of acidic acid
 Since the total volume need to be 50ml, we have
Va+Vb = 50ml

From question 1, we have [base]/[acid] = 5.012 rearrange the equation we get [base] = 5.012*[acid]. This mean that for every 1 ml of acid, we add 5.012 ml of the conjugate base, since the molarity is the same.

now we have 2 equations and 2 unknowns
Va +Vb = 50   and Vb = 5.012Va

assuming you can do the algebra part, we should get Va = 8.32 ml and Vb = 41.86 ml

hope this helps,
Laser

Thank you so much!! I understood exactly how you answered that and I've learnt something So thanks again!