1) 45.0 mL of 0.125 M benzoic acid C6H3COOH is titrated with 0.25 M NaOH. Ka of the acid is 6.46 x 10-5. Calculate the pH of the solutions at the following points of the
titration. Indicate whether the solution is buffer or not at these points.
a. After adding 15 mL of NaOH.
I

answer: you could do some invislibe ink.

explanation: i will tell you how to if u answer my question.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

58750.08068 is the answer i think