Question

Determine the ph of a buffer solution comprised of a 1.41 m hf and 0.623 m naf. the ka of hf = 7.20 x 10-4

Answer 1

The pH of the buffer solution is 2.78.

Step-by-step explanation:

Concentration of salt = 0.623 M

Concentration of acid = 1.41 M

Dissociation constant of acid =
`K_a=7.20* 10^(-4)`

The pH of the buffer solution is given by Henderson-Hasselbalch equation:

`pH=pK_a+\log ([salt])/([acid])`

`pH=-\log[K_a]+\log (salt)/(acid)`

`pH=-\log[7.20* 10^(-4)]+\log(0.623 M)/(1.41 M)`

pH = 2.78

The pH of the buffer solution is 2.78.

Answer 2

we are going to use H-H equation:

PH = Pka + log[conjugate base]/[weak acid]

when we have here the conjugate base is NaF

and the weak acid is HF.

so, we are going to use the Ka value to get the value of Pka.

when Pka = -logKa

= -log7.2 X 10^-4

= 3.14

then we will substitution on H-H equation:

∴PH = 3.14 + log (0.623)/(1.41) = 2.99