Answer:
0.068 moles of sodium hydroxide would have to be added to 150 mL of a 0.483 M acetic acid solution
Step-by-step explanation:
As we know
pH = pKa + log [salt]/[acid]
4.480= 4.74 + log [salt]/[acid]
log [salt]/[]acid] = -0.06
[salt]/[acid] = 0.87
moles acid = 0.150 L x 0.483 mol/L = 0.07245 moles acid
x/0.07245-x = 0.87
X- 0.07245 X = 0.87 *0.07245
X = 0.068 moles