Determine the pH of a 0.100 M solution of hypobromous acid, HBrO, at 298K.
The Ka-2.06 x 10-9

HBrO ⇌ H+ + BrO-1

The pH of the 0.100 M solution of hydobromous acid HBrO is 4.843

Explanation:

Here we have the reaction given as follows;

HBrO ⇄ H⁺ + BrO⁻¹

Therefore;

                   HBrO ⇄ H⁺ + BrO⁻¹

Initial concentration of HBrO = 0.100 M decomposes partly to form x moles of each of H⁺ and BrO⁻¹. That is

                               HBrO          ⇄ H⁺ + BrO⁻¹

Initial                        0.1                     0       0

Change                     -x                     +x     +x

Equilibrium           0.1 - x                   x        x

Hence;

(0.1 - x) × 2.06×10⁻⁹ = x²

x² + 2.06×10⁻⁹·x - 0.1 = 0

Factorizing gives;

(x + 1.4354×10⁻⁵)(x - 1.4352×10⁻⁵) = 0

Therefore, x  = 1.4352×10⁻⁵ M or -1.4354×10⁻⁵ M

We take the positive value as x is the concentration of the ions in the solution;

From the above [H⁺] = 1.4352×10⁻⁵ M

pH = -log[H⁺]  = -log(1.4352×10⁻⁵) = 4.843

The pH of the 0.100 M solution of hydobromous acid HBrO at 298 K where the Ka = 2.06×10⁻⁹ = 4.843.

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