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 ![\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
pH = 4.74 + Log ![\frac{[0.5]}{[0.5]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0.5%5D%7D%7B%5B0.5%5D%7D)
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 ![\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
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
Catalysts speed up a reaction so it would be the reaction rate increases.
Most ions tend to be D. Salts. Hope this helps!
The chemical name of Hc2h3o2 is Acetic Acid.
An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly eleastic and in which there are no intermolecular attractive forces. One can visualize it as a collection of perfectly hard spheres which collide but which otherwise do not interact with each other.
Happy to help