Question

Suppose 10.0 mL of 2.00 MNaOH is added to (a) 0.780 L of pure water and (b) 0.780 L of a buffer solution that is 0.682 Min butanoic acid (HC4H7O2) and 0.674 Min butanoate ion (C4H7O2–). Calculate the pH of (a) and (b) before and after the addition of the NaOH. Assume volumes are additive. (Ka, HC4H7O2= 1.5 × 10-5)

Answer 1

Answer:

a) pH will be 12.398

b) pH will be 4.82.

Explanation:

a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles

The total volume after addition of pure water = 0.780+0.01 = 0.79 L

The new concentration of /NaOH will be:

$molarity=\frac{molesofsolute}{volumeofsolution}=\frac{0.02}{0.79}=0.025M$

the [OH⁻] = 0.025

pOH = -log [OH⁻] = 1.602

pH = 14 -pOH = 12.398

b) The buffer has butanoic acid and butanoate ion.

i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:

$pH=pKa+log\frac{[salt]}{[acid]}$

pKa=$-logKa=-log(1.5X10^{-5})=4.82$

ii) on addition of base the pH will increase.