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The change in pH of a 1.00 L of a buffered solution preparing by mixing 0.50 M acetic acid (Ka = 1.8 x 10^-5) and 0.50 M sodium acetate when 0.010 mole of NaOH is added is 4.75
when the same amount 0.010 mole of NaOH was added to 1.00 L of water the pH = 12
Explanation:
given that:
concentration of acetic acid = 0.50 M
Concentration of base sodium acetate = 0.50 M
ka = 1.8 x 10^-5)
pka = -log [ka]
pka = 4.74
From Henderson-Hasselbalch Equation:
pH = pKa + log
pH = 4.74 + Log
pH = 4.74 + 0
pH = 4.74
Number of moles of NaOH = 0.010 moles
volume 1 litre
molarity = 0.010 M
Moles of acetic acid and sodium acetate before addition of NaOH
FORMULA USED:
molarity =
acetic acid,
0.5 = number of moles
0.5 is the number of moles of sodium acetate.
number of moles of NaOH 0.010 moles
NaOH reacts in 1:1 molar ratio with acetic acid so
number of moles in acetic acid = 0.5 - 0.010 = 0.49
number of moles in sodium acetate = 0.5 +0.010 = 0.51
new pH
pH = pKa + log
pH= 4.74 + log[0.51] - log[0.49]
pH= 4.75
PH of NaOH of 0.01 M (BASE)
pOH = -Log[0.01]
pOH = 2
pH can be calculated as
14= pH +pOH
pH= 14-2
pH = 12