What is the ph of an aqueous solution made by combining 35.81 ml of a 0.4364 m sodium acetate with 38.23 ml of a 0.3183 m solution of acetic acid to which 3.524 ml of a 0.0588 m solution of naoh was added?

pH will be 4.86.

Explanation:

On mixing sodium acetate (a salt of acetic acid) with acetic acid we get buffer solution.

Let us calculate the moles of each species taken:

Moles of sodium acetate = molarity X volume(L)

                                        = 0.4364 X 35.81 X 10⁻³= 0.0156 mol

the moles of acetic acid = molarity X volume(L)

                                        = 0.3183 X 38.23 X 10⁻³= 0.0121 mol

The moles of NaOH added = molarity X volume(L)

                                        = 0.0588 X 3.524 X 10⁻³= 0.000207 mol

The base added will react with the acetic acid producing equal moles of sodium acetate

The moles of acetic acid reacted = moles of base added = 0.000207 mol

The moles of acetic acid left = 0.0121-0.000207 = 0.0119 mol

The moles of sodium acetate formed = 0.000207 mol

The new moles of sodium acetate = 0.0156+0.000207=0.0158 mol

The pH of buffer is calculated using Henderson Hassalbalch's equation

pKa of acetic acid = 4.74

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