I’m not sure how to do this. sorry for cutting off the question at first.

Both those acids are strong acids, so it's a titration with a strong acid and a strong base, creating a neutralization reaction.

a) The pH before any NaOH has been added would just be the -log[HCL]. -log[.1] is equal to 1.

b) First, figure out how many mmol of HCl you have. x/25=.1 would be 2.5 mmol of HCl. There's x/10=.1 of NaOH, so you have 1 mmol of NaOH. 1 mmol of the HCl reacts away, so you have 1.5 mmol of HCl in 35 mL. 1.5/35 = .042 M. -log(.042) = 1.38.

c) In a titration with a strong acid and a strong base, the pH at the equivalence point is always equal to 7.

d) Now you have 10 mL beyond the equivalence point. We need to figure out how much was added to get to the equivalence point so we can calculate the concentration of NaOH to find the pOH and then pH. Well, since they have equal concentrations, it would be 25 mL of NaOH, plus the 10 we just added. So we have 35 mL + 25 mL of the HCl, which is 60 mL total. If we had 35 mL of NaOH originally, that would be 3.5 mmol of NaOH (x/35=.1). 3.5/60 is .0583. -log.0583  is 1.23. pH = 14 - pOH. 14 - 1.23 = 12.77

it is the most commonly accepted one

i think that the answer is c