Which of the following mixtures would result in a buffered solution when 1.0 l of each of the two solutions are mixed?
a. 0.2 m hno3 and 0.4 m nano3
b. 0.2 mhno3 and 0.4 m hf
c. 0.2 m hno3 and 0.4 m naf
d. 0.2 m hno3 and 0.4 m naoh

c. 0.2 M HNO₃ and 0.4 M NaF .

Explanation:

A buffer is defined as the mixture of a weak acid with its conjugate base or a weak base with its conjugate acid.

A weak acid or weak base are defined as an acid or base that partially dissociates in aqueous solution. in contrast, a strong acid or base are acids or bases that is dissociated completely in water.

Thus:

a. 0,2M HNO₃ and 0.4 M NaNO₃. This is a mixture of a strong acid with its conjugate base. IS NOT a buffer.

b. 0.2 M HNO₃ and 0.4 M HF . This is a mixture of two strong acids. IS NOT a buffer.

c. 0.2 M HNO₃ and 0.4 M NaF . NaF is the conjugate base of a weak acid as HF is.

The reaction of HNO₃ with NaF is:

HNO₃ + NaF → HF + NaNO₃

That means that in solution you will have a weak acid (HF) with its conjugate base (NaF). Thus, this mixture IS a buffer.

d. 0.2 M HNO₃ and 0.4 M NaOH. This is the mixture of a strong acid with a strong base, thus, this IS NOT a buffer.

I hope it helps!

space exploration has led to the development and technological improvement of many everyday items. (if this answer is wrong i will find the right answer) : )

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole