What additional volume of 10.0 m hcl would be needed to exhaust the remaining capacity of the buffer after the reaction described in part b? express your answer in milliliters using two significant figures?

A buffer of pH = 7.79 is prepared using Tris HCl and NaOH. Moles of Tris HCl = 31.52 g x (1 mole / 157.59 g) = 0.20 moles.

Moles of NaOH = Molarity x volume = 10.0 M x 6.7 mL / 10^3 L = 0.067 moles

Tris – HCl + OH^- -> Tris + H_2O + Cl^-

Setup the reaction table as follows :

Tris – HCl + OH^- -> Tris + H_2O + Cl^-

0.200           0.067                                            Initial

-0.067          -0.067    +0.067                            Change

0.1333         0              0.067                            Final

The buffer is diluted to ! L and half the solution is taken. So half the moles of tris – HCl and tris are present.

Moles of Tris – HCl = 0.133 / 2 = 0.0665 moles

Moles of tris = 0/067 / 2 0.0335 moles

To this solution, 0.0150 mol of H+ are added. H+ reacts with tris giving Tris – HCl. So moles of tris – HCl = 0.0665 + 0.0150 = 0.0815 moles

Moles of tris = 0.0335 – 0.015 = 0.018 moles

When all the 0.018 moles of Tris are consumed, the remaining capacity of the buffer is exhausted.

So moles of HCl to be added = 0.018 moles.

Volume of Hcl = moles /molarity = 0.018 moles / 10.0 M = 0.0018 L = 1.8 mL

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