Question

Just answers.

A. A buffer solution that is 0.431 M in HClO and 0.431 M in NaClO has a pH of 7.46.

(1) The addition of 0.01 mol of H₃O⁺ to 1.0 L of this buffer would cause the pH to _________(increase slightly, increase by 2 units,decrease slightly,decrease by 2 units,not change.)

(2) The capacity of this buffer for added OH^- could be increased by the addition of 0.147 mol  _________(of the weak acid,of the salt.)

B. How many grams of solid sodium fluoride should be added to 2.00 L of a 5.23×10^-2M hydrofluoric acid solution to prepare a buffer with a pH of 3.816 ?

grams sodium fluoride =  g.

C.How many grams of solid ammonium bromide should be added to 2.00 L of a 0.287 M ammonia solution to prepare a buffer with a pH of 8.740 ?  

grams ammonium bromide =

Answer 1

Three seperate questions are there. First two questions are solved here. kindly post other questions separately.

Q A : Using henderson equation; pH of a buffer solution is :

given that the buffer solution is 0.431 M in HClO and 0.431 M in NaClO, so at this point pH = pKa,

(1) The addition of 0.01 mol of H3O+ to 1.0 L of this buffer, that 0.01 M, will cause the conjugate base, NaClO to completely consume this added H3O+, and its concentration will slightly decrease, hence, from the henderson equation, log term will decrease. Since pka is constant, the overall pH will slightly decrease.

So, correct fill up is pH decreses slightly

(2) On similar grounds, if capacity of this buffer for added OH- has to be increased, than we need extra amount of weak acid, HClO to completely consume the added OH-. Thus weak acid has to be added.

So, correct fill up is of the weak acid .

Q B) pKa = 3.1 of HF,

The buffer would contain weak acid HF and its conjugate base NaF

Agin using the henderson equation; pH of a buffer solution is :

pH= 3.816

So, pH - pKa =   0.716

                     = log ([NaF]/[HF])

So.

[NaF]/[HF] = 100.716 = 5.20

And, [NaF] = 5.20 x [HF]

Since [HF] = 5.23×10-2 M, [NaF] = 5.20 x 5.23×10-2 M = 2.72 ×10-1 M

so mole of solid NaF in 2.00 L of solution = (2.72 ×10-1 M ) x (2.00 L) = 0.544 mol

Molar mass of NaF = 42 g/mol

therefore mass in g of NaF = ( 0.544 mol) x42 g/mol = 22.8 g