How many moles of KOH would you need to add to 1.65 L of buffer (comprised of 0.35 M CH3COOH and 0.35 M CH3COONa) to reach a pH of 4.849? Assume no volume change upon addition of base.
Initial amounts:
mol of acid = MV = 1.65*0.35 = 0.5775 mol of acid
mol of conjugate base = MV = 1.65*0.35 = 0.5775 mol of base
if we need pH = 4.849
apply buffer equation
pH = pKa + log(CH3COONa/CH3COOH )
4.849 = 4.75 + log(CH3COONa/CH3COOH )
(CH3COONa/CH3COOH ) = 10^(4.849-4.75) = 1.256
(CH3COONa/CH3COOH ) =1.256
CH3COONa = 1.256*CH3COOH
therefore... we must add KOH
mol of KOH required = "x"
after adding KOH,
CH3COOH + KOH = K + CH3COO-
CH3COOH decreaes, CH3COO- increases
mol of acid = 0.5775 - x
mol of conjugate base = 1.65*0.35 = 0.5775 + x
now.. .we know the ratio
CH3COONa = 1.256*CH3COOH
(0.5775 - x)1.256 = 0.5775 + x
0.5775 *1.256 - 1.256 x = 0.5775 + x
(0.5775 *1.256 - 0.5775 ) = 2.256 x
x = (0.5775 *1.256 - 0.5775 ) / 2.256
x = 0.0655
mol of KOH required = 0.0655
Get Answers For Free
Most questions answered within 1 hours.