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:
[tex]molarity=\frac{molesofsolute}{volumeofsolution}=\frac{0.02}{0.79}=0.025M[/tex]
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:
[tex]pH=pKa+log\frac{[salt]}{[acid]}[/tex]
pKa=[tex]-logKa=-log(1.5X10^{-5})=4.82[/tex]
ii) on addition of base the pH will increase.
