A certain buffer solution contains equal concentration of X- and textHX . The Ka for textHX is 10- 8 . The textpH of the buffer is

Question

A certain buffer solution contains equal concentration of X- and textHX . The Ka for textHX is 10- 8 . The textpH of the buffer is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The Henderson equation is pH=pKa+log ([C o n j u g a t e textbase]/[A c i d]) =-log Ka+log â¡ ([X-]/[H X]) Hence, [.HX].=[X-] =-log (.10- 8.)+log â¡ 1 or pH = 8.0.

Q. A certain buffer solution contains equal concentration of $X^{-}$ and $HX$ . The $K_{a}$ for $HX$ is $1 0^{- 8}$ . The $pH$ of the buffer is

The Henderson equation is

$p H = p K_{a} + l o g \frac{[ C o nj ug a t e base ]}{[ A c i d ]}$

$= - l o g K_{a} + l o g \frac{[ X ^{-} ]}{[ H X ]}$

Hence, $[H X] = [X^{-}]$

$= - l o g (1 0^{- 8}) + l o g 1$

or pH = 8.0.

Q. A certain buffer solution contains equal concentration of X− and HX . The Ka​ for HX is 10−8 . The pH of the buffer is

The Henderson equation is pH=pKa​+log[Acid][Conjugatebase]​ =−logKa​+log⁡[HX][X−]​ Hence, [HX]=[X−] =−log(10−8)+log⁡1 or pH = 8.0.

Q. A certain buffer solution contains equal concentration of $X^{-}$ and $HX$ . The $K_{a}$ for $HX$ is $1 0^{- 8}$ . The $pH$ of the buffer is

The Henderson equation is

$p H = p K_{a} + l o g \frac{[ C o nj ug a t e base ]}{[ A c i d ]}$

$= - l o g K_{a} + l o g \frac{[ X ^{-} ]}{[ H X ]}$

Hence, $[H X] = [X^{-}]$

$= - l o g (1 0^{- 8}) + l o g 1$

or pH = 8.0.