Question

Calculate the [H+] in a solution that is 0.803 M in NaX and 0.677 M in HX given that the Ka of HX is 8.64 ⋅ 10 − 7 8.64⋅10-7. Report your answer in scientific notation to 3 sig figs.

Answer 1

The concentration of hydrogen ion in the solution was
`7.24* 10^(-7)`.

Step-by-step explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

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

`pH=pK_a+\log(([NaX])/([HX]))`

We are given:

`K_a` = Acid dissociation constant of HX =
`8.64* 10^(-7)`

`pK_a=-\log[K_a]=-\log[8.64* 10^(-7)]=6.06`

`[salt]=[NaX]=0.803 M`

`[acid]=[HX]=0.677 M`

pH = ?

Putting values in above equation, we get:

`pH=6.06+(0.803 M)/(0.677 M)`

pH = 6.14

`pH=-\log[H^+]`

`6.14=-\log[H^+]`

`[H^+]=10^(-6.14)=7.24* 10^(-7)`

The concentration of hydrogen ion in the solution was
`7.24* 10^(-7)`.