One mole methane combusts to form one mole CO2 and 2 moles H2O
I am pretty sure it is d if i am wrong please tell me
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:

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]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
pKa=
ii) on addition of base the pH will increase.