A) 0.10 M formic acid pKa = 3.75
I used the pKa value and did the -log to get the Ka value.
Ka = -log pKa = 10^-3.75 = 1.78x10-4
I have just realised here, I have used 1x1014, which is for working with bases?? Without that...
OH- = √(M x Ka) = √(0.10 x 1.78x10-4) = 4.22x10-3
pOH = -log(OH-) = -log(4.22x10-3) = 2.4
pH = 14 - pOH = 11.6.
Oh dear, I think I have just confused myself. I will have a look at that wesbite.