Explanation:
It is given that molarity of acetic acid = 0.0100 M
Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer
Moles of acetic acid = 0.0100 M × 1.00 L
= 0.0100 mol
Similarly, moles of acetate = molarity of sodium acetat × volume of buffer
= 0.100 mol
When
is added, it will convert acetate to acetic acid.
Hence, new moles acetic acid = (initial moles acetic acid) + (moles
)
= 0.0100 mol + x
New moles of sodium acetate = (initial moles acetate) - (moles
)
= 0.100 mol - x
According to Henderson - Hasselbalch equation,
pH = ![pK_{a} + log\frac{[conjugate base]}{[weak acid]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bweak%20acid%5D%7D)
pH = 
4.95 = 4.75 + 
= 4.95 - 4.75
= 0.20
= antilog (0.20)
= 1.6
Hence, x = 0.032555 mol
Therefore, moles of
= 0.032555 mol
volume of
= 
= 
= 0.0032555 L
or, = 3.25 (as 1 L = 1000 mL)
Thus, we can conclude that volume of
added is 3.26 mL.